THE  MATHEMATICS 


OF 


APPLIED  ELECTKICITY 


A  PRACTICAL  MATHEMATICS 


BY 


ERNEST  H.  KOCH,  Jr. 

Instructor  in  Mathematics,  School  of  Science  and  Technology, 
Pratt  Institute,  Brooklyn,  N.  Y. 


THE  KEY  TO  EVERY  MAN  IS  HIS  THOUGHT' 


FIRST   EDITION 
FIRST  THOUSAND 


NEW  YORK 

JOHN   WILEY  <fc   SONS 

London:  CHAPMAN  &  HALL,  Limited 
1912 


Engineering 
Library 


Copyright,  1912 

BY 

ERNEST  H.  KOCH,  Jr. 


SCIENTIFIC    PRESS 

ROBERT    DRUMMOND   AND    COMPANY 

BROOKLYN,    N.  Y. 


QA37 
K7 

ENGIN. 
LIBRARY 


PREFACE 


This  text  was  developed  primarily  for  the  mathematics 
instruction  of  second  year  students  in  applied  electricity 
courses  at  Pratt  Institute.  It  is  intended  to  follow  a  year 
of  drill  in  the  essential  elements  of  algebra,  plane  geometry 
and  plane  trigonometry  and  also  in  the  elementary  prin- 
ciples of  mechanics,  heat  and  electricity. 

The  introductory  chapter  has  been  written  to  meet  the 
requirements  of  other  young  men  who  are  occupied  in  various 
electrical  industries  and  who  find  their  progress  retarded 
in  electrical  matters  owing  to  their  deficiency  in  the  knowl- 
edge of  elementary  practical  mathematics.  The  text  will 
prove  a  very  helpful  aid  as  a  short  practical  course  in  Trade, 
Industrial  and  Technical  High  Schools,  and  in  Apprentice- 
ship courses. 

The  purpose  of  practical  mathematics  is  to  bring  to  the 
student's  attention  the  underlying  basic  structure  of  the 
relations  of  the  elements  of  electrical  phenomena  so  that 
he  may  formulate  them,  interpret  them  physically  and 
work  with  accuracy  and  facility  in  making  numeric  and 
graphic  computation.  When  correlated  with  a  practical, 
industrial  or  theoretic  course  in  electricity,  practical  mathe- 
matics gives  the  student  a  better  appreciation  of  his  major 
subject  and  assures  him  an  intellectual  penetration  into 
secrets  of  nature  and  the  utilization  of  her  powers.  Instruc- 
tion from  the  mimeographed  form  of  this  text  has  been 
conducted  for  a  number  of  years  and  has  demonstrated  its 

253761  m 


iv  PREFACE 

usefulness  as  a  powerful  and  interesting  presentation  which 
has  promoted  the  student's  judgment,  fidelity,  poise,  com- 
mon sense  and  tact  in  his  applied  work  and  in  subsequent 
success  in  foremanship,  superintendence  and  in  other  exec- 
utive and  administrative  positions.  The  author  aims  to 
present  the  text  according  to  the  modern  principles  of 
education  which  makes  the  text  not  only  teachable  but  also 
self-instructive.  Although  the  work  is  graded  and  arranged 
in  logical  sequence  it  is  so  written  that  it  may  be  presented 
in  any  order  irrespective  of  chapter  divisions.  There  is 
considerable  material  of  a  more  advanced  character  which 
may  be  omitted  on  first  assignment  and  used  later  for 
review  or  in  further  preparation  for  more  advanced  study. 

One  aim  has  been  to  use  simple  diction,  good  English, 
and  carefully  phrased  statements  which  shall  assist  the 
student  in  becoming  familiar  with  technical  terms  and 
guide  him  in  expressing  his  observations  in  his  own  language. 
Abstract  mathematic  theory  has  been  avoided  as  non- 
essential since  the  author  realizes  that  any  so-called  utili- 
tarian or  practical  book  cannot  fail  to  provide  the  so-called 
cultural  aspects  of  education.  The  text  contains  a  limited 
but  an  adequate  amount  of  information  concerning  electrical 
phenomena  so  as  to  minimize  the  necessary  references 
to  supplementary  electrical  texts.  Problems  and  examples 
are  both  designated  under  the  abbreviation  of  exercise — Ex. 

For  convenience  the  text  is  divided  into  three  parts,  viz. 
I.  The  Transformation  and  Interpretation  of  Formulas. 
Direct  Current  Problems.  II.  The  Graphs  of  Formulas 
and  the  Formulation  of  Graphs.  III.  Vectors  and  Vector 
Diagrams,  Alternating  Current  Problems. 

I  have  collected  problems  during  a  number  of  years 
from  many  sources,  making  it  impossible  to  give  specific 
acknowledgment.  I  am  indebted  to  my  colleagues  for 
many  suggestions  and  express  my  thanks  and  appreciation 
to  Mr.  H.  W.  Marsh,  Head  of  the  Department  of  Mathe- 
matics,  to  Mr.   S.   S.   Edmands,  Director  of  the    School 


PREFACE  v 

of  Science  and  Technology,  Pratt  Institute,  and  to  Mr. 
A.  L.  Williston,  Principal  of  Wentworth  Institute,  who 
inspired  the  creation  of  the  text.  I  am  indebted  to  many 
of  my  former  students,  more  particularly  to  Messrs.  C.  H. 
Meeker,  H.  A.  Keteham  and  J.  G.  Brown,  for  valuable 
assistance  in  preparing  plates  and  diagrams.  It  gives  me 
great  pleasure  to  commend  the  engravers,  printers  and 
publishers  for  their  careful  and  conscientious  labors  in 
producing  this  book. 

E:  H.  Koch,  Jr. 

Newark,  N.  J.,  May,  1912.  ' 


TABLE  OF  CONTENTS 


PART  I 

THE  TRANSFORMATION  AND  INTERPRETATION  OF 
FORMULAS.    D.C.   PROBLEMS. 

CHAPTER  I 
Fundamental  Operations 

PAGE 

Numbers,  Magnitude,  Measurement  —  The  Decimal  System 
— Addition,  the  Use  of  Letters  as  Symbols  of  Abbrevia- 
tion— Subtraction,  Parentheses — Multiplication,  Factors — 
Properties  of  Simple  Geometric  Constructions — Axioms  and 
Their  Applications — Summary  of  Theorems  on  Lines  and 
Angles  —  Demonstration  —  Triangles,  Quadrilaterals  and 
Polygons — Areas  of  Figures — Variation,  Ratio,  Proportion — 
Theorems  on  Proport'ons — Projection — Trigonometric  Func- 
tions— The  Use  of  Trigonometric  Tables — Functions  of 
Angles  of  Any  Quadrant — The  Right  Triangle — Square  Root 
Operations  upon  Fractions — Laws  of  Numbers — The  Oblique 
Triangle — Logarithms — The  Slide  Rule — Trigonometric  and 
Logarithmic  Tables   1 

CHAPTER  II 
The  Transformation  of  Formulas 
The  Formula  —  The  Notation  —  The   Data  —  The  Transforma- 
tion— The  Solution — The  Interpretation — The  Analysis  of  a 
Formula — Transformation  of  Formulas 124 

CHAPTER  III 
Interpretation  of  Formulas 131 


viii  TABLE  OF  CONTENTS 

CHAPTER  IV 
The  Formulation  of  Problems 

PAGE 

The  Ampere-hour 133 

CHAPTER  V 

Variation  Problems  Applied  to  the  Electric  Circuit 

Variation  of  Resistance  with  Length  and  Area  —  Mnemonics  of 

Resistance — Square  and  Circular  Mil 136 

CHAPTER  VI 
The  Algebra  of  a  Simple  Electric  Circuit 145 

CHAPTER  VII 

The  Applications  of  Ohm's  Law 

Current,  E.M.F. — A  Resistor,  Resistance — Drop  of  Potential — 
Power  —  Conductance  —  Temperature  Coefficient  —  British 
Thermal  Unit,  Pound  and  Small  Calorie  —  Efficiency  of 
Transformation  and  Transmission — Net  Work  in  an  Elec- 
tric Circuit,  Kirchhoff's  First  Law — Kirchhoff's  Second  Law — 
Back  Pressure 151 

CHAPTER  VIII 

Efficiency  of  Generators  and  Motors 

Efficiency — Electrical  Efficiency — Commercial  or  Net  Efficiency 

— Gross  Efficiency .' 179 

CHAPTER  IX 

The  Algebra  of  the  Magnetic  Circuit 

Comparison  of  Magnetic  and  Electric  Circuit  Relations — Mag- 
netizing Force  —  Flux  Density  —  Reluctance  —  Permeance 
— The  Field  within  a  Coil,  a  Solenoid — Ampere-turns — 
Reluctance  of  a  Compound  Circuit — Total  Ampere-turns — 
Leakage  Permeance — The  Flux  in  a  Generator — The  Energy 


TABLE  OF  CONTENTS  ix 

PAGE 

Loss  due  to  Hysteresis — The  Energy  Loss  due  to  Eddy  Cur- 
rents— The  Average  E.M.F. — The  E.M.F.,  Generated  in  a 
D.C.  Armature — Torque — Flux  Density  in  Armature 182 

CHAPTER  X 

Winding  Calculations 

The  Dimensions  of  Insulated  Wire — Square  Winding — Stagger 
or  Imbedded  Winding — Number  of  Turns — Diameter  of  Wire 
in  Terms  of  Resistance  and  Winding  Volumes — Volumes 
of  Winding  Space 197 

CHAPTER  XI 
Formulas  of  Mensuration 

Units  of  Area  and  Volume — Area  of  a  Parallelogram,  Triangle, 
Trapezoid,  Regular  Polygon,  Circle,  Annulus  or  Circular 
Ring,  Ellipse  —  Squaring  the  Circle  —  Rectification  of  a 
Circular  Arc  —  Approximate  Perimeter  of  an  Ellipse  —  The 
Sector  and  Segment  of  a  Circle — Area  of  Segment  of  Ellipse, 
Parabola — Approximate  Length  of  a  Parabola  Segment  of 
an  Hyperbola — Area  of  an  Irregular  Figure,  by  Planimeter, 
Squared  Paper,  Weighing,  Mean  Ordinate  Rule,  Mid  Ordinate 
Rule,  Trapezoidal  Rule,  Simpson's  One-third  and  Three- 
eighths  Rule,  Durand's  Rule,  Weddle's  Rule — The  Cycloid 
—The  Helix  — The  Spiral— The  Conical  Spiral  — Unit  of 
Volume  —  Cube  —  Rectangular  Prism  — ■  Triangular  Prism — 
Oblique  Prism — Quadrangular,  Pentagonal,  Hexagonal,  Right 
and  Oblique  Prisms — Total  Surface  of  a  Prismatic  Figure 
— The  Volume  of  a  Pyramid — Volume  of  a  Cone — Plane 
Sections — Frustum  of  a  Pyramid  or  Cone — Volume  of  a 
Prismoid — Ungula  of  a  Solid — Cylindric  Surface,  Surface 
of  a  Cone — Figures  of  Revolution — Guldinus'  Theorems, 
Solid  Rectangular  and  Circular  Rings — Comparison  of  Cylin- 
der, Cone  and  Sphere — Segment  of  Sphere,  Spherical  Zone, 
Oblate  and  Prolate  Spheroids — Volume  of  Segment  of  Sphe- 
roid, Paraboloid,  Hyperboloid — Regular  Solids — Volumes  of 
Irregular  Solids  —  Binomial  Theorem  —  Approximations 
and  Errors — Applications  of  Pythagorean  Theorem 202 


x  TABLE  OF  CONTENTS 

CHAPTER  XII 
The  Quadratic  Equation 

PAGE 

Degree  of  an  Equation  —  Solving  a  Quadratic  Equation  by- 
Factoring — Solving  a  Quadratic  Equation  by  Completing 
the  Square — The  Graphic  Construction  of  the  Roots  of  a  Quad- 
ratic Equation — The  Removal  of  the  Radical  from  an  Equa- 
tion— Indicated  Operations  and  Inverse  Operations — Homo- 
geneous Equations — Parenthetical  Quantities 247 

CHAPTER  XIII 

The  Elements  of  the  Strength  of  Materials 

Stress  per  Unit  Area  —  The  Strength  of  any  Material  —  A 
Beam — Lever  Arm,  Moment — Bending  Moment — A  Canti- 
lever— Center  of  Gravity,  Centroid — Moment  of  Inertia, 
Radius  of  Gyration — Biquadratic  Inches — Neutral  Axis, 
Resisting  Moment — I-beam,  T-beam,  Channel — Comparative 
Strength  of  Beams  of  Equal  Length  and  Equal  Cross-section 
— Stiffness,  Deflection — Polar  Moment  of  Inertia — Combined 
Bending  and  Twisting — Pulley  Diameter,  Armature  Bearings 
— Riveted  Joints  and  Their  Efficiency 259 


PART  II 

THE  GRAPHS  ARE   FORMULAS  AND  THE  FORMULATION 
OF  GRAPHS 

CHAPTER  XIV 

The  Use  of  Squared  Paper 

Graphic  Methods,  Data,  Cross-section  Paper — Plotting  Tabulated 
Data  —  Transformation  of  the  Axes  —  Plotting  of  a  Third 
Variable — Plotting  of  Formulas — The  Graphic  Representa- 
tion of  Ohm's  Law 279 


TABLE  OF  CONTENTS  xi 

CHAPTER  XV 
Linear  Graphs 

PAGE 

Preparation  of  Tables  and  Plates  —  Slope  —  Dependent  and 
Independent  Variables — Characterization  of  Linear  Graphs- 
Degree  of  an  Equation — The  Proof  that  a  Linear  Graph 
Represents  a  Simple  Equation — Intercepts — Writing  the 
Equation  of  a  Given  Straight  Line — Graphic  Representation 
of  Simultaneous  Equations 300 

CHAPTER  XVI 

Non-Linear  Algebraic  Equations 

A  Non-Linear  Equation,  Algebraic  Equations — Parabolic  Curves 
—  Family  of  Standard  Parabolas  —  Families  of  Curves 
—Odd-Odd,  Odd-Even,  Even-Odd  Parabolas— The  Composi- 
tion of  Curves  by  Addition  and  Subtraction — Hyperbolic 
Curves — Powers,  Roots,  Reciprocals — Other  Non-Linear 
Equations,  Power  Series — Conic  Sections,  Conic  Test — 
The  Meaning  of  the  xy  Term — The  Solution  of  Non-Linear 
Simultaneous  Equations 316 

CHAPTER  XVII 

Eccentricity  of  Conics 

Analytic  Definition  of  a  Conic  —  Eccentricity  —  Construction 
of  Conics  with  Straightedge  and  Compass — Properties  of 
Parabolas  and  Hyperbolas 342 

CHAPTER  XVIII 

The  Formulation  of  Graphs 

Writing  the  Equation  of  a  Curve  —  Linear  Forms  of  Non- 
Linear  Equations — Logarithmic  Cross-section  Paper — The 
Plotting  of  Curves  on  Logarithmic  Paper — Special  Forms 
of  Cross-section  Paper  for  Linear  Plotting  of  Non-Linear  Equa- 
tions— Exponential  Equations — The  Exponential  Family  of 
Curves — Exponential  Table 348 


xii  TABLE  OF  CONTENTS 

CHAPTER  XIX 

The  Use  of  Polar  Paper 

PAGE 

Polar  Coordinates  —  Polar  Paper  —  Plotting  on  Polar  Paper 
—Polar  Equations — Polar  Equation  of  Conies,  Circle — The 
Logarithmic  Spiral — Roulettes,  Cycloid,  Epicycloid,  Hypocy- 
cloid — Evolutes  and  Involutes — The  Trisection  of  an  Angle, 
Conchoid — The  Multisection  of  an  Angle,  Chordel 367 

CHAPTER  XX 

Solving  Formulas  by  Charts 

Graphic  Solution  by  Charts — Linear  Coordinates — Chart  Solution 
of  Quadratic  and  Cubic  Equations — The  Change  of  Chart 
Scales — The  Reduction  of  a  Cubic — The  Charting  of  Linear 
Forms,  Reciprocal  Forms  and  Composite  Forms — Chart  of 
Charts : 383 

CHAPTER   XXI 

Measurement  of  Angles 

Angle  Designation — Radian  of  Arc,  Radian  of  Angle — Verification 
of  Trigonometric  Equations — Functions  of  Algebraic  Sums 
of  Angles — Functions  of  Half  Angles — Tangent  Functions 
—  Exponential  and  Trigonometric  Series  —  Hyperbolic 
Functions 393 

CHAPTER  XXII 

Harmonic  Motion 

Simple  Harmonic  Motion  —  The  Sine  Curve,  a  Record  of 
Simple  Harmonic  Motion — Plotting  the  Fundamental  Sine 
Curve — Angular  Velocity,  Period,  Frequency — Comparison 
of  Sine  Curves  of  Different  Frequency — The  Amplitude  of  a 
Sine  Curve — Angle  of  Phase — The  Resultant  of  Two  Simple 
Harmonic  Motions — The  Product  of  Two  Sine  Curves,  Power 
Curves — The  Curve  of  Damped  or  Decaying  Oscillation — 
The  Graphic  Solution  of  Inverse  Functions 401 


TABLE  OF  CONTENTS  xiii 

CHAPTER  XXIII 

Rates,  Derivatives  and  Integrals 

PAGE 

Speed  —  Velocity  —  Slope  —  Scale  of  Velocity  —  Initial  Average 
and  Final  Velocity  —  Linear  and  Angular  Velocity  —  The 
Derivative — The  Graphic  Determination  of  the  Derivative 
of  a  Curve — Primitives  and  Derivatives — Differentiation  of 
Formulas — The  Proof  of  the  Law  of  the  Derivative  of  a  Power 
— The  Operation  of  Obtaining  a  Derivative — Integration  of  a 
Power,  Exponential  and  Logarithmic  Forms — An  Abridged 
Table  of  Integrals — Graphic  Integration — The  Graphic  Inte- 
gration of  a  Linear  Form — Graphic  Double  Integration — 1  he 
Scales  of  Integral  Curves — The  Integration  Formula  for  the 
Area  of  a  Curve — The  Area  of  the  Sine  Loop — Mean  Effec- 
tive Value  of  an  Harmonic  E.M.F.,  or  Current — The  Moment 
of  Inertia  Formula  for  a  Structural  Section — The  Static  Mo- 
ment Formula  for  any  Figure — Graphic  Determination  of 
Static  Moments,  Moments  of  Inertia,  and  Centers  of  Grav- 
ity of  Sections — Nomenclature  and  Interpretation  of  Dif- 
ferential and  Integral  Expressions 430 


CHAPTER  XXIV 

Experimental  Curves 

The  Effect  of  the  Variation  of  Load  and  Terminal  Voltage 
on  the  Operation  of  a  Shunt  Motor — Characteristic  Curves  of 
a  Compound  Generator — Characteristic  Curves  of  a  Series 
Generator — The  Changes  in  the  Current,  Voltage  and  Density 
of  the  Electrolyte  of  a  Storage  Battery  during  Discharge — 
The  Efficiency  Test  of  a  Shunt  Motor — Comparison  of  the 
Starting  Torque  of  Shunt  and  Series  Motors — The  Perform- 
ance of  a  Rotary  Converter — Electrical  Measuring  Instru- 
ments— Siemens  Dynamometer,  Thomson  Ammeter,  Queen 
Ammeter,  Whitney D. C.  Voltmeter,  Weston  Instruments.  .  .  .   483 


xiv  TABLE  OF  CONTENTS 

PART   III 

VECTORS  AND  VECTOR  DIAGRAMS.    A.C.    PROBLEMS 

CHAPTER  XXV 
Inductance  and  Capacity  - 

PAGE 

Counter  E.M.F.  of  Self-induction  —  Unit  of  Inductance  — 
Inductive  Reactance — Inductance  Formulas — Mutual  Induc- 
tance— Practical  Values  of  Inductances — Capacity — Capacity 
of  Condensers — Table  of  Dielectric  Constants — Practical 
Values  of  Capacities — Formulas  for  Calculating  Capacities — 
Parallel  Connection  of  Condensers — Series  Connection  of 
Condensers — Condensers  Connected  in  Series  Parallel-Ar- 
rangement —  The  Energy  Equation  of  an  A.C.  Circuit — 
Reactance 504 

CHAPTER  XXVI 

Elementary  Operations  op  Vector  Analysis 

Vectors — Vector  Representation — Equal  Vectors — The  Addition 
of  Vectors — Concurrent  Forces — The  Vectorial  Representa- 
tion of  Currents  and  E.M.Fs 524 

CHAPTER  XXVII 

Vector  Algebra 

Symbolic   Representation   of    a    Vector — Vector    Notation — The 

Magnitude  and  Inclination  of  a  Vector 536 

CHAPTER  XXVIII 
The  Graphic  Solution  of  A.C.  Problems 

Circuits  having  Resistance  and  Inductance  —  Power  Factor  — 
Circuits  having   Resistance  and  Capacity — Circuits  having 


TABLE  OF  CONTENTS  xv 

PAGE 

Inductance,   Capacity  and  Resistance — General  Expressions 
for  an  A.C.  Circuit 539 

CHAPTER  XXIX 

Series  Circuits 

The  symbols  for  K,  X,  Z  —  Standard  Frequencies  —  Problems 
Solved  Graphically  and  Numerically — Resonance — Problems 
Solved  by  Adding  Impedances  546 

CHAPTER  XXX 

Parallel  Circuits 

Graphic    and    Numeric    Treatment    of    Problems     Relating    to 

Parallel  Circuits — Equivalent  Impedance — Circle  Diagram. .  .   559 

CHAPTER  XXXI 

Series  Parallel  Circuits 570 

CHAPTER  XXXII 
Alternating  Current  Problems 
A-  and  ^-Connections — Transmission  Lines 573 

CHAPTER  XXXIII 

The  Algebra  of  a  Transformer 

The  Transformer  —  Equivalent  Resistance  and  Reactance  — 
Losses  and  Efficiency — Design  of  Core — Connections — Auto- 
transformer  590 

CHAPTER  XXXIV 

The  Circle  Diagram 

Variable  Elements  in  an  A.C.  Circuit — Fundamental  Principles 
in  an  A.C.   Circuit — The  Horizontal  Auxiliary  Circle — The 


xvi  TABLE  OF  CONTENTS 

PAGK 

Construction  of  the  Phase  Angle — The  Vertical  Auxiliary 
Circle — The  Eccentric  Circle — Multiple  Auxiliary  Circles — 
Reactance  and  Resistance  in  Series  with  a  Constant  Voltage 
Circuit — Parallel  Circuits  with  Constant  Applied  Voltage — 
Relabeling  the  Lines  of  a  Circle  Diagram  —  The  Deter- 
mination of  the  Equivalent  Resistance  and  Equivalent 
Reactance — Parallel  Circuits  whose  Constant  Elements  are 
Unlike — The  Circle  Diagram  for  a  Constant  Current  Circuit 
— Circuits  having  Constant  Resistance  and  Constant  Induct- 
ance with  Variable  Frequency — Multiple  Circuit  Circle 
Diagrams 601 


APPENDIX 

Greek  Alphabet — Instructions  for  Preparing  and  Reporting  Daily 

Work 631 


PRACTICAL    MATHEMATICS 


CHAPTER  I 
FUNDAMENTAL   OPERATIONS 

1.  Numbers,  Magnitude,  Measurement.  Numbers  are 
combinations  of  nine  figures — 1,  2,  3,  4,  5,  6,  7,  8,  9,  called 
digits,  the  letter  0,  called  naught,  cipher  or  zero,  and  the 
period,  called  the  decimal  point. 

These  numeric  characters  are  written  either  singly  or 
together  without  commas  separating  them.  They  express 
rank,  order,  greatness,  vastness,  extent,  size,  intensity, 
activity,  strength  or  significance  of  quantities. 

A  scale  is  a  measuring  stick,  instrument,  or  device 
upon  which  there  is  a  series  of  graduated  lines  numbered 
in  sequence.  Any  division  or  part  of  a  scale  may  be  used 
as  a  unit  of  measure  and  thereby  extend  the  range  of 
measurement. 

The  magnitude  of  a  quantity  is  the  number  of  units 
which  it  contains.  The  magnitude  of  a  5-lb.  article  is  5 
and  not  5  lbs.  If  the  weight  were  expressed  in  ounces, 
its  magnitude  would  be  eighty. 

Ex.  1.  In  the  following  measurements  use  the  most 
convenient  devices  at  hand,  such  as  a  yardstick,  foot-rule,  tape- 
measure,  store  scale.  In  as  manv  ways  as  possible  express  the 
magnitude  by 


2  PRACTICAL  MATHEMATICS 

(1)  Measuring  the  length,  breadth  and  height  of  your  room. 

(2)  Measuring  the  length,  breadth  and  thickness  of  a  door.       \ 

(3)  Measuring  a  table,  desk,  box,  quart  can,  ash  can,  oil  can. 

(4)  Measuring  the  parts  of  a  motor,  generator,  engine,  shafting. 

(5)  Measuring  the  sides  and  perimeters  of  regular  and  irregular 

sections  of  materials  provided  by  the  instructor. 

(6)  Weighing   a   quart   of  water,  sand,    cement,  loose   earth, 

packed  earth,  100'  No.  18  copper  wire. 

(7)  Measuring  the  length,  breadth,  and  thickness  and  also  by 

weighing  a  copper  bus-bar,  bars  of  steel,  wrought  and 
cast  iron. 

(8)  Measuring  and  weighing  a  brick,  cement  building  block. 

(9)  Describe  at  least  ten  other  scales  not  used  in  the  above 

examples. 

The  observation  from  the  above  list  of  examples  is  that 
magnitudes  express  the  results  of  measurements  or  comparisons. 
Magnitude  is  a  number  which  states  how  many  and  not 
what  kind  of  units  the  object  contains.  The  unit  of  measure 
must  possess  the  same  quality,  characteristic,  i.e.,  denomi- 
nation of  the  object  measured.  The  unit  of  measure,  as  its 
name  implies,  has  a  magnitude  of  one. 

2.  The  Decimal  System.  In  the  decimal  notation,  all 
numbers  are  based  on  a  system  of  ten,  making  it  possible 
to  write  a  number  with  comparatively  few  figures.  The 
annexing  of  zeros  after  a  number  multiplies  by  powers 
of  ten.  A  number  followed  by  one  0  is  multiplied  by  10. 
A  number  followed  by  00  is  multiplied  by  100.  A  number 
followed  by  000  is  multiplied  by  1000,  and  so  on.  We 
then  have  the  following  summary,  assuming  the  original 
number  to  be  95623 : 

95623  =  95623X1 
956230  =  95623X10 
9562300  =  95623  X 100  =  956230  X 10 
95623000  =  95623  X 1000  =  9562300  X 10 

It  is  observed  that  the  multiple  value  of  each  figure  depends 
on  its  position  in  a  given  number.     Numbers  from  1  ...  9 


FUNDAMENTAL  OPERATIONS  3 

inclusive  require  one  figure.  To  write  a  number  ten  times 
as  great  move  the  figure  one  place  to  the  left  and  fill  in  its 
former  'position  with  a  zero. 

Suppose  the  process  is  reversed.  We  begin  with  the 
number  95623000.  To  write  a  number  one-tenth  of  this 
value  remove  one  zero  on  the  right  and  move  each  figure 
one  place  to  the  right.  If  this  process  be  continued  the 
number  will  be  divided  by  ten  in  each  stage.  Although 
we  may  continue  to  move  the  figures  to  the  right  we  shall 
have  exhausted  all  the  zeros  in  three  stages,  i.e.,  we  have 
reached  the  number  95623.  To  write  a  number  one-tenth 
of  95623,  move  the  figures  one  place  to  the  right  and  insert 
a  decimal  point  between  2  and  3.  To  write  a  number  one- 
tenth  of  the  last  result,  move  the  figures  one  place  to  the 
right  and  shift  the  decimal  point  one  place  to  the  left,  and 
so  on.     This  is  illustrated  below: 

95623000 

9562300  m  iV  of  95623000 

956230   =iV  of  9562300 

95623   =  A  of  956230 

9562.3  =tV  of  95623 

956.23=  A  of  9562.3 

Fractions  whose  denominators  are  powers  of  ten  are 
united     into     a     decimal.     Thus     1+JL+JL+-J- 

Q  Q  fj  f> 

= — + — H — --\ -  =  .3976.     Each    numerator    takes    its 

10     102     103     10* 

position  in  the  decimal  answer  corresponding  to  the  power 

of  10  in  its  denominator. 

The  number  956.23  means  9  times   100  and  5  times 

10  and  6  units  and  2  tenths  and  3  hundredths.     In  other 

2  3 

words  956.23  means  900  and  50  and  6  and  —  and  — — ,  or 


4  PEACTICAL  MATHEMATICS 

2  3 

to  900  add  50  add  6  add  —  add  — .     The  symbol  for  add 

is  (+)  read  plus,  so  the  above  may  be  written,  900+50+6 
+.2+. 03,  or,  rearranging  in  a  vertical  column,  we  have 

900 
50 
6 
.2 
.03 


956.23 


Therefore  956.23  is  a  condensed  form  which  saves  space 
and  the  writing  of  zeros.  It  is  read  nine  fifty-six  point 
twenty-three,  or  nine  five  six  point  two  three. 

3.  Addition.  The  addition  of  numbers  is  a  process  of 
condensation  or  uniting  of  figures  in  the  same  column, 
file  or  row.  An  excess  over  nine  in  any  column  is  counted 
into  the  next  column  to  the  left. 


:.  2. 

(1) 

(2) 

(3) 

(4) 

(5) 

(6) 

(7) 

(8) 

12 

37 

48 

52 

73 

94 

50 

66 

37 

64 

78 

32 

90 

33 

62 

98 

45 

76 

21 

45 

67 

78 

32 

54 

68 

43 

64 

38 

23 

79 

22 

65 

77 

11 

13 

6 

60 

5 

4 

23 

29 

46 

7 

8 

23 

54 

59 

31 

column  j    38 
sums       1 23 


sum  268 

The  process  is  identically  the  same  when  the  numbers  are 
arranged  horizontally,  12+37+45  +  68+77+29=268.  In  the 
latter  case  the  figures  are  united  in  the  order  of  their  numeric 
position.     The  symbol  (  = )  means  equals  or  is  equal  to. 


FUNDAMENTAL  OPERATIONS  5 

It  is  of  primary  importance  for  the  computer  to  be  able 
to  recognize  instantly  the  sum  of  two  integers.  In  the 
following  arrangement  of  two  reversed  rows  of  integers 
the  student  should  recognize  that  each  column  sums  10. 


1 

2 

3 

4 

5 

6 

7 

8 

9 

9 

8 

7 

6 

5 

4 

3 

2 

1 

Ex.  3.  Write  the  integers  in  any  mixed  order,  then  state 
or  write  immediately  the  complementary  number.  Numbers  are 
complementary  when  their  sum  is  10. 

Ex.  4.  Rearrange  the  digits  in  double  rows  so  that  their 
sums  will  gives  nines,  eights,  sevens,  sixes,  fives,  fours,  threes, 
twos,  ones.  It  will  be  observed  that  some  of  these  combinations 
are  repetitions.     There  should  be  45  different  combinations.     Why? 

Ex.  5.  What  is  the  sum  of  the  nine  digits?  What  is 
the  sum  of  the  next  ten  numbers?  What  is  the  sum  of  the  ten 
numbers  following?  What  is  the  increase  for  any  group  of  ten 
numbers  following  a  set  of  ten  numbers? 

The  addition  of  numbers  should  be  checked  by  adding  the 
columns  from  the  top  downward  as  well  as  from  the  bottom 
upward.  Where  the  numbers  contain  many  figures,  the  sums  of 
each  column  should  be  written  below  the  columns,  and  finally 
the  total  sum  should  be  written  at  the  bottom  as  shown  in  the 
following  example.  The  marks  *,  t,  °,  \  to  the  right  of  the  figures 
indicate  that  those  similarly  marked  were  grouped  in  the  summing. 

The  sum  of  the  figures  in  the  units  column  is  66,  which  is  the 
same  as  60  and  6.  Therefore  we  write  6  in  the  units  column 
and  6  under  the  tens  column.  The  sum  of  the  figures  in  the 
tens  column  is  18,  which  means  that  we  have  18  times  10,  or  180, 
or  100  and  80.  Therefore  we  write  8  under  the  tens  column 
and  1  under  the  hundreds  column.  Proceeding  in  like  manner 
with  the  other  columns,  we  obtain  80  for  the  sum  of  the  hundreds 
column,  which  means  80  times  100,  or  8000.  Therefore  we  write 
8  under  the  thousands  column  and  zero  or  nothing  under  the 
hundreds  column.  The  sum  of  the  thousands  column  is  45,  which 
means  45  times  1000,  or  45,000,  or  40,000  and  5000.  Therefore 
we  write  4  under  the  ten-thousands  column  and  5  under  the 
thousands  column.  The  sum  of  the  ten-thousands  column  is  9. 
Therefore  we  write  9  under  the  ten-thousands  column. 

The  figures  set  down  under  the  columns  are  added.  Their 
sum  gives  the  sum  total,  which  is  143,246.     Setting  down  the 


6 


PRACTICAL  MATHEMATICS 


figures  under  the  columns  enables  us  to  check  the  work  more 
rapidly. 

excess  over  9=6 

3 

6 

3 

8 

4 

0 

0 

6 

6-  9  5*  4'  6 

1  3°  3  0     1'  8 

1  6°  6  0    2  6 


3  1°  4 

1  5  7 
7*  8 
3*  9 

2  7f  8  1 
1  3f  9  0 

4'  5 


6* 

8t 
4f 

7 

8t 

9 

5* 

9* 

3' 

4' 

r 

2 


6    6 

excess  over  9=3 

56  excess  over  9 

the  sums 

1  8 

0 

of  each       8 

0 

8 

column    4  5 

0 

9 

0 

sum     14  3    2  4    6  excess  over  9=2      11  excess  over  9=2 


•J2 

I 


&     O     2      <D 


■"O  -^> 


la 

p£3     -+-J     +i       rG     +3 


«  2 


The  sum  total  is  a  short  way  of  stating  that  we  have  altogether 
100,000,  40,000,  3,000,  200,  40,  and  6.  In  words  we  have  one 
hundred-thousands  plus  four  ten-thousands  plus  three  thousands 
plus  two  hundreds  plus  four  tens  plus  six  units.  Therefore 
100000  +40000  +3000  +200  +40  +6  =  143246. 

An  additional  check  is  obtained  by  casting  out  the  nines. 
Write  the  excess  over  9  from  the  sum  of  the  digits  in  each  row 
and  also  from  the  sums  of  excesses.  The  final  excesses  2  are  the 
same  when  the  addition  has  been  performed  without  error. 

The  Use  of  Letters  as  Symbols  of  Abbreviations.  In 
mathematics  work  the  alphabet  serves  not  only  to  express 


FUNDAMENTAL  OPERATIONS  7 

our  ideas  in  words,  but  also  to  symbolize  objects  by  abbre- 
viating their  names  by  a  contracted  form  or  by  a  single 
letter.  Thus  amperes  may  be  abbreviated  by  amp;  volts 
by  v;  kilo-watts  by  k.w.;  electromotive  force  by  E.M.F. 
or  by  E;  intensity  of  current  by  I;  feet  by  ft.;  inches 
by  in.;  miles'by  mi.;  circular  mils  by  CM.,  etc. 

The  engineering  alphabet  is  suggested  because  of  its 
simplicity.  The  letters  are  constructed  of  thin  straight  lines 
and  arcs  of  circles.  The  letters  are  equally  broad  and 
high  with  the  exception  of  the  I,  M,  W.  The  sides  of  the  M 
are  vertical,  whereas  the  sides  of  the  W  are  inclined.  Arcs 
occur  in  the  letters  B,  C,  D,  G,  J,  0,  P,  Q,  R,  S,  U. 

Ex.  6.  Obtain  a  piece  of  paper  ruled  in  squares  and  upon 
it  prepare  an  alphabet  of  capital  and  also  an  alphabet  of  lower 
case  letters.     Follow  this  by  the  nine  digits.     Repeat  for  practice. 

Whenever  the  letters  of  the  alphabet  are  used  for 
abbreviations  we  should  be  consistent  in  the  use  of  either 
CAPS  or  1.  c,  and  should  not  change  from  one  style  to  the 
other  unless  the  conditions  of  the  problem  require  it. 

A  succession  of  figures  or  letters  connected  by  +  signs 
means  that  the  figures  or  letters  are  to  be  united  by  addition. 
Thus  n+5n+3n  means  that  a  number  which  is  represented 
by  the  letter  n  is  to  be  added  to  five  times  the  same  number 
n  and  in  addition  three  times  the  same  number  n  is  to  be 
added  to  the  preceding  amount.  The  total  result  or  sum 
will  be  9n  or  (1  +  5+3)  n,  or  (one + five + three)  times  n. 
Suppose  n  stands  for  321,  then  5n  =  5  times  321  =  1605, 
and  3n  =  3  times  321  =  963.  Therefore  instead  of  n+5n+3n 
we  may  write  321  +  1605  +  963,  which  equals  2889  =  9 
times  321  =9n. 

Observation.  The  use  of  a  letter  to  abbreviate  a  number 
saves  time,  space,  and  unnecessary  multiplication.  The 
number  written  in  front  of  n  is  its  multiplier  and  is  called 
the  coefficient  of  n.  The  coefficient  expresses  the  number  of 
times  the  attached  letter  is  used  in  addition.     5n  is  therefore 


8  PKACTICAL  MATHEMATICS 

a  contraction  of  n-\-n-}-n-\-n-{-n,  and  Zn  is  a  contraction 
of  n-\-n-\-n.  In  like  manner  9n  is  a  double  contraction, 
because  9n  is  first  of  all  a  contraction  of  n+5n+3n,  which 
is  in  turn  a  contraction  of  n-\-n-\-n-\-n-\-n-\-n-\-n-\-n-\-n.  In 
addition  the  sum  of  the  coefficients  of  any  letter  is  the  coeffi- 
cient of  the  same  letter  in  the  answer. 

Ex.  7.    Perform  the  following  indicated  additions: 

(1)  26+56+76+36+116  =?6=how  many  6's? 

(2)  7h+4h+h+6h  +  10h  =28?  =twenty-eight  of  what? 

(3)  5k  +5k  +5k  +5k  +5k  =  ??  =how  many  of  what? 

(4)  2E  +SE  +E+8E=?E  =  how  many  E's? 

(5)  3a+6v+5a+v+3a+7v  =  ?a+?v=how  many  a's  plus  how 

many  tf's? 

If  the  example  contains  different  letters,  these  must  be 
summed  separately,  as  different  letters  represent  different 
things  and  therefore  cannot  be  united.  This  is  exactly  the 
the  same  kind  of  a  statement  we  would  make  if  we  were 
asked  to  render  an  inventory  of  electrical  apparatus  in  a 
stockroom  or  in  a  shop  or  laboratory.  There  would  be 
one  sum  obtained  by  adding  the  number  of  lamps,  another 
sum  would  be  obtained  by  adding  the  number  of  switches, 
and  other  sums  to  correspond  to  the  different  types  of 
equipment.  Suppose  we  were  given  the  addition  of  20s + 
502p+36c+57s+31p  +  92c+32s+173p+19c,  the  answer 
would  be  109s+706p+147c.  Since  s,  p,  and  c  respectively 
represent  different  pieces  of  apparatus.  In  expressing  the 
sum  we  state  how  many  there  are  of  each  kind,  i.e.,  how 
many  c's,  p's,  s's. 

Observation.  Addition  can  be  made  only  of  quantities 
of  the  same  kind  or  of  the  same  denomination.  Therefore  in 
a  series  of  additions  we  unite  quantities  or  terms  with  like 
letters. 


FUNDAMENTAL  OPERATIONS  9 

Ex.  8.      Perform  the  following  indicated  additions: 

(1)  36+2c+5d+26+4c+3.2d+6+c+5.7d+2.5c 

(2)  8.1a+x+Zx+gc+—  a+-x+2.5a+2 

The  student  may  find  it  convenient  to  arrange  the 
work  in  columns,  putting  like  quantities  in  the  same  col- 
umn as  shown  below: 


36+2c     +  5d 

3'6+2     ;c+  5 
2       4             3.2 

d 

63+2      +1 

6 

26     4c          3. 2d 

c2+4      +1     +2.5 

8.5 

b      c          5.7d 

1 

1            5.7 

2.5 

d.5+3.2  +5.7 

13.9 

2.5c 

66+8.5c+13.9d 

66+8.5c+13.9d 

6 

6+8.5c+13.9 

d 

(3)  312+6Z+523&  +2|*+319.2&+139 

(4)  bmi  +6yd +7ft  +3rai  +308/*  +25yd  +45?/d  +2.5/Z 

4.  Subtraction.  When  we  wish  to  subtract  one  quan- 
tity from  another  quantity,  a  minus  sign  (  — )  is  written 
preceding  the  subtrahend.  Thus  6h  —  4:h  means  from  the 
minuend  6h  subtract  4/i  the  subtrahend.  The  remainder 
is  2h,  which  is  the  excess  of  6h  over  4/i. 

If  a  series  of  quantities  are  to  be  united  so  as  to  involve 
both  addition  and  subtraction  the  operations  may  be  per- 
formed in  any  order. 

(5)  7a+3a-5a+4a-2a 

Example  (5)  means  to  la  add  3a,  then  subtract  5a,  then 
add  4a,  and  then  subtract  2a.  Then  the  remainder  or 
that  which  is  left  over  or  the  excess  is  7a.  The  analysis 
of  the  procedure  is  as  follows : 

Since  7a+3a  =  10a,  then  7a+3a  — 5a+4a— 2a  becomes  10a 

—  5a+4a— 2a. 
Since  10a  —  5a  =  5a,  then  10a  —  5a  +  4a  —  2a  becomes  5a 

+4a-2a. 
Since  5a+4a  =  9a,  then  5a+4a  — 2a,  becomes  9a  — 2a. 
Since  9a  —  2a  =  7a,  then  9a  — 2a  becomes  7a. 


10  PRACTICAL  MATHEMATICS 

Therefore 

7a+3a-5a+4a-2a=7a. 

From  the  above  it  will  be  observed  that  there  were  a 
total  of  14  a's  added  and  a  total  of  7  a's  subtracted  and 
again  14a — la  gives  the  excess  7a.  In  the  following  examples 
we  have  the  same  quantities  and  operations  to  deal  with 
as  in  ex.  (5),  except  that  the  order  of  the  quantities  has 
been  changed.  Show  that  the  remainder  is  the  same  in 
each  case.  When  a  sign  is  omitted  before  the  first  term 
the  plus  sign  is  understood  as  intended. 

(6)  7a+3a+4a-5a-2a  (9)  3a+7a-5a-2a+4a 

(7)  3a+4a-5a+7a-2a  (10)  3a-5a+4a+7a-2a 

(8)  3a+4a-2a-5a+7a  (11)   -2a+3a+4a-5a+7a 
(12)  add  the  columns  formed  by  examples  (6)  .  .  .  (11). 

Observation.  Subtraction  is  the  reverse  operation  to  addi- 
tion. When  the  minuend  is  greater  than  the  subtrahend  the 
excess  is  positive,  and  when  the  minuend  is  less  than  the 
subtrahend  then  the  excess  is  negative  and  indicates  a  deficiency. 
The  latter  would  correspond  to  a  debt  or  liability  in  a  system 
of  accounts,  whereas  the  former  would  correspond  to  a  credit 
or  asset.  A  negative  amount  of  any  quantity  will  always 
cancel  a  like  positive  amount  of  the  same  quantity.  It  is  for 
this  reason  that  the  excess  always  takes  the  sign  of  the 
uncanceled  amount  in  any  series  of  additions  and  subtractions. 

In  an  election  contest  -\-A  may  represent  the  number 
of  votes  received  for  a  candidate  and  —B  the  number  of 
votes  against  a  candidate.  Suppose  the  favorable  votes 
are  10568  and  the  unfavorable  votes  are  8717.  Then  the 
plurality  of  the  candidate  over  his  opponent  is  an  excess 
of  A  over  B,  or  A -5  =  10568-8717  =  1851  plurality.  In 
other  words  the  candidate  has  received  10568  (+  votes) 
and  8717  (-  votes). 

The  minus  sign  preceding  a  quantity  may  be  regarded 
in  two  ways.     We  may  interpret  it  as  a  symbol    of   oper- 


FUNDAMENTAL  OPERATIONS  11 

ation,  i.e.,  a  command  to  subtract  or  remove  or  take  away 
the  thing  which  follows  it.  We  may  also  regard  the  thing 
which  follows  the  minus  sign  as  a  deficiency  or  something 
which  is  stored  away  or  passive  or  inactive.  It  would 
therefore  have  a  contrary  sense  to  a  like  positive  quantity 
which  might  represent  something  being  used  or  operative 
or  active. 

In  a  tug  of  war  there  are  pulls  in  both  directions.  If 
we  call  those  acting  to  the  right  positive  pulls  and  those 
acting  to  the  left  negative  pulls,  then  the  sum  total  of  all 
the  effects  is  zero  when  the  +  and  —  pulls  balance. 
There  is  a  +  or  —  excess  depending  upon  the  preponderance 
of  the  +  or  —  pulls  respectively. 

During  part  of  the  day  when  the  demand  for  power 
from  a  station  is  low  the  plant  may  be  operated  efficiently 
by  storing  its  excess  of  energy  in  a  bank  of  batteries. 
Suppose  a  500-k.w.  machine  is  running  at  full  load,  and 
only  300  k.w.  of  this  amount  is  supplied  to  a  lamp  circuit, 
then  the  balance  200  k.w.  is  inactive  and  is  stored  in  a 
series  of  storage  batteries.     Then, 

Total  power = power  used  in  lamps+power  stored  in  batteries. 

(A)  "500  k.w.  =300  k.w.+200  k.w. 
This  may  be  rewritten: 

Total  power  =  useful  or  active  power— inactive  power. 

(B)  500  k.w.  =  +300  k.w. -(-200  k.w.) 

The  +  sign  before  300  k.w.  indicates -f- sense  or  active  power. 
The  —  sign  before  200  k.w.  indicates  —  sense  or  inactive  power. 
The   —   sign  before  the  parenthesis   (  )  means  stored  or 

removed  or  subtracted.     (A)  and  (B)  are  two  different 

ways  of  expressing  the  same  idea. 

Observation.  A  double  negative  sign  preceding  a  quantity 
restores  the  quantity  to  a  positive  sense.  Therefore  every 
example  in  subtraction  is  changed  to  one  of  addition  by  chang- 
ing the  sign  of  the  subtrahend,  and  adding  the  quantities. 


12  PRACTICAL  MATHEMATICS 

Ex.  13.    What  is  the  meaning  of  6a;  -  {7x  -3a;)? 

6z  is  the  minuend  and  (7x—Sx)  is  the  subtrahend.  The 
(7x—Sx)  is  preceded  by  a  —  sign  and  is  therefore  to  be  sub- 
tracted from  6x.  The  parenthetical  quantity  alone  implies  the 
subtraction  of  Sx  from  7x  and  reduces  to  4x.  Therefore  the 
(7x  —Sx)  may  be  replaced  by  its  equal  Ax  and  the  original  example 
may  be  restated  as  Qx—Ax.  The  result  or  excess  is  2x.  The 
same  result  would  have  been  obtained  if  the  original  expression 
had  been  altered  by  removing  the  (  )  and  reversing  the  signs  of 
all  the  quantities  contained  within  it.     Thus : 

6z  -  (7x  -Sx)  =  6x  -7x  +Sx  =  2x. 

These  forms  are  also  equivalent  to 

Qx  +  {  -7x  +Sx)  =Qx  +  ( -Ax) =Qx  -\x  =2x. 

Observation.  When  a  +  sign  precedes  a  (  )  there  is  no 
need  to  retain  the  (  ).  When  a  —  sign  precedes  a  ( )  the  ( ) 
can  be  removed,  only  providing  we  change  the  signs  preceding 
each  term  inside  the  ( ). 

The  terms  are  the  letters  or  numbers  or  combinations 
of  both  which  are  separated  distinctly  by  +,  — ,  or  =  signs. 

A  parenthesis  ( )  is  a  grouping,  bonding  or  aggregating 
symbol.  If  an  expression  requires  more  than  one  group 
of  terms  it  is  customary  to  use  different  forms  of  bonding 
symbols  to  limit  or  define  the  extent  of  each  group.  The 
following  bonding  symbols  are  equivalent  to  the  ( )  and 
may  replace  one  another: 


bracket;     {     }  brace;  vinculum. 


6x-(7x-3x)=6x-[7x-3x]=6x-{7x-3x}  =Qx-7x-3x 
=  Qx—4:X  =  2x. 

Although  the  vinculum  is  usually  written  over  the 
terms  which  are  to  be  bonded,  it  is  also  recognized  as  the 
solidus  or  the  dividing  line  which  is  written  under  the 
terms  of  a  numerator  to  show  that  all  the  terms  above  it 
are  to  be  operated  upon  by  a  common  divisor.     These 


FUNDAMENTAL  OPERATIONS  13 

grouping  symbols  are  especially  useful  when  it  is  desirable 
to  indicate  that  a  number  of  terms  are  to  be  subjected  to 
a  like  operation. 

3a  (5  — 66)  means  that  both  5  and  6  b  are  to  be  multiplied 
by  3a,  giving  the  result  15a— 186a. 

2x(y— 3a— Qb)=2xy— Qax— 12bx  means  that  each  one  of 
the  three  terms  on  the  right  of  the  equality  sign  contains  a 
common  factor  or  part.  The  parenthesis  indicates  not  only 
that  the  2x  is  common  to  all  the  terms,  but  it  also  shows  the 
remaining  constituent  parts  or  factors  of  each  term. 

^A+B+C=^A+B+C 

means  that  the  sum  of  A,  B,  and  C  are  collectively  and 
not  singly  subject  to  the  root  symbol.  If  A  =6,  £  =  9, 
and  C  =  10,  then 

v/A+B+C  =  v6+9+l6=^/25  =  5. 


Ex.  14.  What  is  the  distinction  between  VX2  +R2  and  \/X2  + 
R2  ?    Illustrate  by  assuming  X  =9  and  R  =4. 

5.  Multiplication.  Numeric  calculations  should  be  made 
in  the  easiest  manner  so  as  to  economize  in  time,  effort, 
and  the  chance  of  mistakes.  Thus  in  multiplying  6305  by 
1453,  we  may  write  either  of  these  numbers  as  the  multiplier 
and  then  use  the  other  number  as  the  multiplicand.  In 
other  words,  6305X1453  =  1453X6305  =  9161165.  The 
latter  order  saves  one  line  of  multiplication  as  follows: 


6305 

1453 

1453 

3X6305 

6305 

18915  = 

7265  = 

5X1453 

31525  = 

5X6305 

0  = 

0X1453 

25220  = 

4X6305 

4359  = 

3X1453 

6305   = 

1X6305 

8718   = 

6X1453 

9161165  =  1453X6305  9161165  =  6305X1453 


14  PKACTICAL  MATHEMATICS 

Since  the  figures  change  only  on  writing  the  partial 
product  we  need  not  write  the  multiplier  and  multiplicand. 
This  method  of  omission  is  illustrated  below:  63.05X14.53 
=  916.1165: 

18915  .7265 

31525  43.59 

25220  878.8 

6305 


916.1165  916.1165 

Observation.  The  multiplier  times  the  multiplicand  equals 
the  product;  also  the  multiplicand  times  the  multiplier  equals 
the  product. 

If  we  abbreviate  the  numbers  6305,  1453,  and  9161165 
by  substituting  the  letters  a,  b,  and  p,  respectively;  times 
by  X  ;  equals  by  =  ;  then  we  may  write  the  product 

6305X1453  =  9161165    as    aXb  =  p, 

1453X6305  =  9161165    as    bXa  =  p. 

Since  letters  are  used  to  abbreviate  quantities  they  are 
used  in  computations  in  exactly  the  same  way  as  the  numbers 
for  which  they  have  been  substituted.  The  letters  of  a 
product  are  usually  written  close  together,  so  that  the 
multiplication  symbol  "X,"  is  omitted  or  contracted  to  a 
point.  It  will  be  understood  that  two  or  more  letters,  or 
letters  and  numbers,  written  close  together  imply  multi- 
plication.    The  factors  are  the  makers  of  a  product. 

aXb  =  a -b  =  ab  =  product  =  p, 

bXa  =  b-a  =  ba  =  product  =  p. 

Factor  times  Factor  =  Product. 

Since  we  obtain  identical  products  whether  we  multiply 
a  by  b  or  multiply  b  by  a,  we  write  accordingly, 

aXb  =  bXa    or    ab  =  ba. 


FUNDAMENTAL  OPERATIONS  15 

Observation.  Things  equal  to  the  same  thing  are  equal  to 
each  other.  This  statement  is  known  as  the  Axiom  of  Equality 
and  is  abbreviated  by  ax=ty. 

An  axiom  is  a  truth  which  the  mind  recognizes  as  self- 
evident  but  which  it  cannot  prove  or  demonstrate.  When 
a  fact  is  at  the  bounds  of  our  most  fundamental  ideas  there 
is  no  expression  more  simple  in  terms  of  which  to  present  it. 

Observation.  In  multiplication  the  order  of  multiplier 
and  multiplicand  does  not  affect  the  result. 

A  product  always  consists  of  two  or  more  parts  or  makers 
called  factors. 

The  computation  of  multiplication  may  be  abbreviated 
whenever  the  arithmetic  product  is  to  be  expressed  with  a 
few  significant  figures.  Sometimes  the  multiplier  is  used 
backward,  as  shown  below: 

Ex.  15.  Multiply  7235  by  1294,  and  express  the  answer  with 
four  significant  figures. 

7235                                   7235 
1294  4912  


29000  7235 

65100  1447 

14470  651 

7235  29 


93620000  9362 

Therefore  7235x1294=9362X1000  approximately  =9362  XlO3. 

Ex.  16.  By  ordinary  and  contracted  methods  determine  the 
following  products  to  four  significant  figures: 

(a)  6305X333  (e)  2819x4567 

(6)  7070X212  (/)   2000X356 

(c)  8532X6729  (g)  3.325x267 

(d)  4. 56 X. 00653  (h)    . 0379 X. 0045 

Ex.  17.  What  is  the  meaning  of  1000  =  10x10x10=10'? 
The  three  written  above  and  to  the  right  of  the  ten  is  called  an 
exponent.    It  indicates  that  1000  is  composed  of  three  equal  "  10  " 


16  PRACTICAL  MATHEMATICS 

factors.    In  other  words,  the  "10"  is  made  use  of  three  times  in 
producing  1000. 

In  the  same  manner  the  product  aaa  which  consists  of  three 
equal  "  a  "  factors  is  abbreviated  by  a3,  or  a  Xa  Xa  =aaa  =a3. 

Observation.  In  multiplication  we  add  exponents  of  like 
factors  and  write  this  sum  over  the  like  factor  in  the  product. 

A  product  of  two  equal  "  b  "  factors,  three  equal  "  z  " 
factors,  and  four  equal  "  m "  factors  may  be  written 
"  bbzzzmmmm"  but  with  the  use  of  exponents,  this  is 
written  62m4z3  or  b2z3m*.  The  first  form  gives  preference 
to  alphabetic  order  and  the  second  form  gives  preference 
to  numeric  order. 

Observation.  An  exponent  serves  to  abbreviate  multipli- 
cation. 

Ex.  18.    630500000  =  6.305  X 100  000  000  -  6.305  X 108.    Why  is 
108  an  abbreviation  for  100  000  000? 
Ex.  19.    Explain  the  following  forms: 

(a)  923  500  000  =9235  X105 
(6)  17,280  X  1000  =  1728X10* 

(c)  8753  X103  =8.753  X108 

(d)  783abc  X  lOabd  =  7.83  X  Wa^cd 
(c)  a2b3c4Xabc2=azbAc* 

(/)   5xy  X2yz  xSxz2  =  30x2y2z3 
(g)  6(a+6)x7(a+6)=42(a+6)2 

The  following  notations  are  often  used  where  it  is  found 
desirable  to  carry  the  contraction  still  further.  A  divisor 
may  be  brought  up  into  the  numerator  as  a  factor,  pro- 
viding the  sign  of  the  factor's  exponent  is  changed  from  a 
positive  to  a  negative  sense,  or  vice  versa. 

(h)   89360000  =  8936  X104  =  893604. 

(i)    0.000001236  =  0.051236  -  0.051236. 

7892 
(J)  78.92  =  7.892X10  =  —— =  7892X10-2. 


FUNDAMENTAL  OPERATIONS  17 

Explain  the  following  forms: 

(m)   0.937  =  9.37  X10_1  =  937  XHT3 

(n)  0.00546  =  5.46  X10"3  =  546  X10-5 

,  ,    3.14X108     ,  ,  .       1         1       1       1 

(0)    ^78T  =  ?  (q)  ^00  =  l0Xl0XI5  =  10"3- 

W  iSXiSXl^-1^X1^X1^"I^-1Crl0- 
Give  four  significant  figures  in  the  answers  of  the  following: 
Examples 

20.  7891X9944. 

21.  7529x3366. 

22.  5278X3246. 

Perform  the  following  by  reversing  the  multiplier: 

23.  6547x6531. 

24.  8054x9601. 

25.  6665XS772. 

When  the  numeric  values  of  the  factors  of  a  product  are 
given  or  known,  they  may  be  substituted  for  the  letters  and 
thereby  enable  the  numeric  value  of  the  product  to  be 
obtained. 

Ex.  26.  Determine  the  numeric  value  of  b2m2z3  by  substitut- 
ing the  following  values:  6=2;  m  =  ll;  2=5.  From  the  data  we 
obtain 

62=22=2X2=4;  ra2=ll2  =  ll  Xll  =121;  z3  =53  =5x5x5=125 

but  b2m2z3  means  b2  times  m2  times  z3,  therefore, 

b2m2z*  =4X121X125  =60500. 


18  PRACTICAL  MATHEMATICS 

The  multiplication  of  4  by  121  may  be  performed  first 
or  the  4  may  be  multiplied  by  125,  or  we  may  proceed  by 
multiplying  121  by  125  first.  The  product  so  obtained  is 
then  multiplied  by  the  remaining  factor.  In  every  case 
the  final  product  is  the  same, 

4X121X125=  484X125  =  60500 

4X125X121  =  5000X121  =  60500 

121X125X4  =  15125  X  4  =  60500 

Do  these  facts  verify  any  previously  observed  law? 

Ex.27.  What  is  the  value  of  m2s3i/4a2,  given  m=2;  s=3; 
y=2.5;  a  =  10? 

Every  example  in  multiplication  should  suggest  the 
best  system  to  be  used  in  performing  the  work  quickly 
and  accurately. 

Ex.  28.  Multiply  62.5  by  25.  Consider  25=—-,  then  the 
example  may  be  written: 

™.    „.     ^r     100    62.5X100    62.5     mM 

62.5X25  =62.5  X  — = z =-r-Xl00 

.4  4  4 

and 

^  =  15.625; 

but 

15.625X100=1562.5; 

,\     62.5X25=1562.5. 

Observation.  When  the  multiplier  is  25,  divide  the  given 
number  by  4  and  multiply  by  100.  The  multiplier  100  shifts 
the  decimal  point  two  places  to  the  right. 

Ex.  29.    Multiply  5.23  by  9.    Consider  9  =  (10  - 1),  therefore, 

5.23x9=5.23(10-1)  =5.23x10-5.23=47.07. 


FUNDAMENTAL  OPERATIONS  19 

Observation.  When  the  multiplier  is  9  shift  the  decimal 
point  one  place  to  the  right  and  then  subtract  the  original 
number. 

Ex.  30.    Multiply  78.91  by  72.36.    We  may  write 

Express  the  answer  as  a  decimal. 

Observation.  The  product  of  several  factors  has  as  many 
decimal  figures  as  the  sum  of  the  several  decimal  figures  in 
the  factors. 

Ex.  31.  Devise  quick  methods  for  performing  the  following 
multiplications: 

(a)  3366X9.449  (c)   72.6x19 

(b)  8.93x27  (d)  25x24 

The  decimal  point,  although  the  smallest  symbol  in 
computation,  is  nevertheless  the  most  important. 

In  dividing  62.5  by  4  the  quotient  should  not  be  written 
156.25,  because  the  result  should  not  be  larger  than  the 
numerator  62.5. 

In  dividing  7.02  by  0.351  the  quotient  should  not  be 
written  2,  because  the  result  should  not  be  smaller  than 
the  numerator  7.02. 

Observation.     A  quotient  is  smaller  than  the  dividend  or 

numerator  when  the  divisor  or  denominator  is  greater  than  one} 

and  is  larger  than  the  dividend  when  the  divisor  is  less  than 

one. 

0.996 

should  not  be  written  9.75,  because  the  numerator 

of  the  fraction  is  smaller  than  the  denominator,  which  means 
that  the  quotient  should  be  less  than  one. 


20 


PEACTICAL  MATHEMATICS 


Observation.  A  proper  fraction  reduces  to  a  value  less 
than  one  and  an  improper  fraction  reduces  to  a  value  greater 
than  one. 

6.  Division.  Division  may  be  performed  in  the  usual 
manner,  or,  where  the  answer  is  required  to  be  given  with 
a  few  significant  figures,  then  the  contracted  form  suggested 
below  will  prove  advantageous. 

Ex.  32.    Divide  6295  by  1453.     This   may  be  written  in  the 

following  equivalent  forms:  6295-^1453  or  6295  :  1453  or 
6295/1453: 


1453 


or 


Dividend     _     ,.  Numerator  . 

=  Quotient  =  — : —   —=  fraction  =  Ratio. 


6295 
5812 

4830 
4359 


Divisor 
Ordinary  Method 
1453  Divisor 


Denominator 


Contracted  Method 


4.33241  Quotient 


4710 
4359 

3510 
2906 


6295 
5812 

483 
436 

47 
44 

3 
3 


1453 


Divisor 


4.332  Quotient 


6040 
5812 

2280 
1453 

827  Remainder 

Romance  Method 
4.332 

1453 


Decimal  Method 


4.332  Quotient 


6295 
4830 
4710 
3510 


6295.000 
5812 

483 
435  9 

47  1 
43  59 

3  51 
2  906 

FUNDAMENTAL  OPERATIONS  21 

Ex.   33.    By  ordinary  and  contracted  methods,  determine  the 
value  of  the  following  fractions  to  four  significant  figures: 

,  x  9999 
(a)1274' 

(6)  7831 


(c) 
(d) 


6103' 

0.9999 
0.1274' 

0.7831 
0.6592' 


w 

3533 
172.3' 

(/) 

3756 
9834' 

(9) 

9.009 

15.06' 

(h) 

0.001293 

CL 

(i)  What  is  the  value  of  —  when  a  =7.03  and  b  =9.52? 
o 

(?)  What  is  the  value  of  —  when  w=2;  y=27:  z  =6? 

z 

Give  an  approximate  mental  estimate  of  the  values  in  the 
following  examples  and  then  determine  the  exact  values  by  cal- 
culation: 

(A;)  327X56.3,  (I)    .002x1.0018x0.996, 

•      1.003  ,  .     1 

(m)0997'  (P)98' 

(n)  T005'  {q)  506' 

1.0023X0.9984  1_ 

(0)     1.0016X0.98  '  (r)  1.009' 

7.  Properties  of  Simple  Geometric  Constructions.  The 
retention  of  more  decimal  figures  than  the  work  warrants 
is  dishonest  and  without  meaning  in  commercial  practice.. 
The  degree  of  accuracy  with  which  we  make  our  measure- 
ments determines  the  number  of  significant  figures  which 
should  be  used. 

Ex.  34.  Use  a  straightedge  and  draw  a  line  about  4.5 
inches.  For  convenience  in  referring  to  this  fine  we  label  it  by 
placing  a  letter    "L  "  above  it.     With  a  12-inch  scale  or  rule 


22  PRACTICAL  MATHEMATICS 

measure  "L"  to  the  nearest  quarter-inch  division.  Express  the 
answer  as  a  mixed  number  and  also  in  terms  of  its  decimal 
equivalent. 

Remeasure  "  L  "  to  the  nearest  eighth-inch  division.  Express 
the  answer  as  a  mixed  number  and  also  as  a  decimal.  How  many 
decimal  figures  are  dependable? 

Remeasure  "  L,"  using  sixteenths  of  an  inch  as  the  unit  of 
measure.  In  expressing  the  answer  be  careful  not  to  use  more 
figures  than  the  work  warrants. 

Ex.  35.  Using  rg  of  an  inch,  the  smallest  subdivision  of  a  12- 
inch  rule,  measure  the  length  of  a  pencil.  Suppose  we  discover  that 
there  are  7  full  inches  and  a  remainder  lying  between  j%  and  yV 
scale  divisions,  but  nearer  the  former.  We  should  be  correct  if 
we  write  the  length  of  the  pencil  =7re,f  =  W5"  =  115  times  the 
unit  A".  Now  7^=7.1875=7.19,  approximately.  The  answer 
is  not  reliable  beyond  7.18,  and  therefore  the  figures  75  should  be 
omitted  and  added  in  as  1  to  8  in  the  second  decimal  place.  This 
is  explained  by  considering  the  fact  that  we  cannot  measure  closer 
than  one-half  of  the  smallest  scale  division.  Therefore  the  degree 
of  accuracy  equals 

^       115  measured  divisions    230 
i  divison  (estimated)        1 

Therefore  the  figure  8  or  9  in  the  second  decimal  place  is  ques- 
tionable, because  an  error  1  too  large  or  1  too  small  in  this  means, 
an  acccuracy  of  one  part  in  700. 

Explain  why  the  answer  may  have  been  written  7.21  and  still 
lie  within  the  limits  of  accuracy. 

Ex.  36.  Three  quantities,  A,  B,  C,  are  measured.  The 
number  of  units  in  ^4=8096,  the  number  of  units  in  5=9.91, 
and  the  number  of  units  in  (7=0.025.  Suppose  an  error  of  1  is 
made  in  the  last  figure  of  each  measurement,  what  is  the  degree 
of  accuracy  in  each  result? 

Observation.  The  degree  of  accuracy  is  not  indicated  by 
the  position  of  the  decimal  point,  but  by  the  number  of  sig- 
nificant figures. 

Ex.  37.  Rule  a  sheet  of  paper  into  squares  as  shown 
on  page  25  by  drawing  horizontal  and  vertical  lines  equally 
spaced  \"  apart.  Leave  a  blank  marginal  space  at  the  four  edges. 
This  is  called  squared  cross-section  paper  and  also  squared  paper. 
The  trade  supplies  such  paper  divided  into  eighths  of  an  inch, 
tenths  of  an  inch,  and  also  into  millimeters. 


FUNDAMENTAL  OPERATIONS 


23 


Fig.  4  shows  the  method  of  using  a  T-square  and  a 
right  triangle  for  constructing  and  testing  parallel  and 
perpendicular  lines.  The  .  T-square  has  its  upper  edge 
planed  true,  i.e.,  straight,  and  is  therefore  a  straightedge. 
The  T-square  is  held  rigidly  to  the  paper  to  prevent  its 
slipping,  while  the  triangle  is  slid  along  the  straightedge. 
The  two  triangles  shown  in  Fig.  5  are  used  in  exactly  the 
same  way  as  the  T-square  and  triangle  shown  in  Fig.  4, 
excepting  that  either  of  the  triangles  may  serve  as  the 
straightedge.  Lines  which  are  parallel  are  marked  in  a 
like  manner  by  single,  double,  or  triple  strokes  or  crosses. 
The  two  right  triangles  are  called  30°-60°  and  45°  respect- 
ively. Test  the  angles  of  the  triangles  and  the  sum  of 
the  angles  of  the  triangles. 


Fig.  1. — A  Protractor. 


Ex.  38A.  Test  the  angles  of  your  squared  paper  with  T-square 
and  right-angled  triangles  and  measure  the  angles  with  a  pro- 
tractor. An  angle-  has  two  sides.  The  point  of  intersection  of 
the  sides  is  called  the  vertex.  To  measure  an  angle  with  a 
protractor,  place  the  center  of  the  latter  at  the  vertex  of  the 
angle  with  the  zero  mark  of  the  graduated  arc  over  one  side  of 
the  angle,  read  the  nearest  degree  of  arc  over  the  other  side  of 
the  angle.  In  Fig.  1  the  angle  AOB  is  measured  by  placing  the 
center  of  the  protractor  at  the  vertex  0  and  0°  of  the  protractor 
on  side  OB,  then  OA  cuts  the  protractor  at  58°.  Therefore 
AOB  =58°. 


24  PRACTICAL  MATHEMATICS 

Lines  (always  refer  to  straight  lines)  are  designated  by 
two  capital  letters  placed  usually  at  the  extremities  of  the 
line  or  by  a  single  lower  case. letter  which  usually  corre- 
sponds to  the  capital  letter  at  the  opposite  vertex.  When 
the  line  forms  part  of  a  figure  (see  Fig.  2)  a  small  arc  is 
often  drawn  at  the  vertex  of  the  angle  with  an  arbitrary 
number  inserted  in  the  arc  instead  of  the  number  of  degrees. 
There  may  be  several  angles  having  an  equal  degree  meas- 
ure, but  these  may  be  distinguished  readily  with  accents 
by  calling  them  angles  1,  1',  1",  V".  If  they  are  unequal, 
it  is  better  to  call  them  more  distinctly  by  angles  1,  2,  3, 
4,  etc.  Arcs  are  usually  designated  by  three  letters  (see 
Fig.  7).  Consult  the  list  of  symbols  and  the  list  of  abbre- 
viations. 

i  Ex.  38B.  Through  the  lower  left-hand  corner  0  of  the  squared 
paper  draw  three  lines  making  angles  9,  10,  11,  respectively,  with 
the  lower  ruled  bounding  line,  see  Fig.  3.  As  near  as  your  eye 
can  judge  without  measuring,  make  these  angles  approximate  to 
30°,  45°,  60°,  respectively.  Now  use  the  protractor  to  measure 
angles  9,  10,  11,  and  also  mark  and  measure  the  angle  11  minus 
angle  10;  angle  11  minus  9;  angle  10  minus  9.  What  is  the 
degree  of  accuracy  in  each  measurement,  assuming  that  we  can 
estimate  to  one-half  of  a  degree? 

Observation.  In  measuring  quantities  with  the  same  unit 
of  measure  the  greatest  magnitude  has  the  greatest  degree  of 
accuracy. 

Two  parallel  lines  lie  in  the  same  plane  and  do  not  meet 
or  intersect  within  a  finite  distance  even  when  produced 
or  extended.  Lines  are  either  parallel  or  non-parallel. 
Name  the  sets  of  parallel  and  non-parallel  lines  on  the 
squared  paper,  which  you  have  constructed.  Why  are 
they  parallel  or  non-parallel? 

Intersecting  vertical  and  horizontal  lines  are  perpen- 
dicular. A  perpendicular  is  a  line  which  forms  a  right 
angle,  i.e.,  a  90°  angle  with  another  line.  Name  the  sets 
of  perpendicular  and  non-perpendicular  lines  on  the  squared 


FUNDAMENTAL  OPERATIONS 


25 


A 

1 

B 

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c 

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K 

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c 

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L 

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Pin-   ± 

XL 

rij 

V 

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Q 

N 

R 

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1 

i 

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vx 

v\. 

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Ss 

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l 

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K J 

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X 

2 

xx 

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X 

2 

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3  2 

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2        p 

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vx. 

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Fig.: 

J 

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tl\ 

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JD 

26  PRACTICAL  MATHEMATICS 

paper,  which  you  have  constructed.     Why  are  they  per- 
pendicular or  non-perpendicular? 

Ex.  39.  Construct  Fig.  6,  which  represents  a  portion  of  the 
squared  paper,  with  two  of  the  horizontal  parallel  lines  AB  and 
CD,  accentuated.  These  are  cut  at  E  and  F  respectively  by 
the  transversal  GH.  It  is  called  a  transversal  because  it  is  a 
straight  line  which  cuts  across  several  other  lines.  These  inter- 
secting lines  form  four  angles,  1,  2,  3,  4,  at  E,  and  four  angles, 
5,  6,  7,  8,  at  F.  If  these  angles  are  measured  and  taken  in  pairs, 
we  find  they  are  either  equal  or  supplementary. 

Angles  whose  sum  is  180°  are  supplementary.  Angles 
whose  sum  is  90°  are  complementary. 

Angles  having  a  common  side  and  common  vertex  and 
two  sides  exterior  are  adjacent. 

Angles  6  and  3,  5  and  2,  8  and  1,  7  and  4,  are  supple- 
mentary.    Verify  by  measurement. 

Angles  1  and  4,  1  and  3,  3  and  2,  2  and  4,  are  supple- 
mentary adjacent.  Why?  Name  all  other  supplementary 
adjacent  angles  in  Fig.  6. 

Vertical  angles  lie  opposite  to  the  intersection  of  two 
lines  and  are  non-adjacent  but  equal.  1  and  2  are  vertical 
angles.     Name  the  other  vertical  angles. 

Interior  angles  lie  within  the  parallels  and  exterior 
angles  lie  outside  the  parallels.  When  angles  lie  on  oppo- 
site sides  of  the  transversal  they  are  called  alternate.  When 
they  lie  on  the  same  side  they  are  called  unilateral  or  direct. 

Angles  2  and  6  are  alternate  interior.  Name  one  other 
pair  of  alternate  interior  angles.  Are  they  equal  or  unequal? 
Why? 

Angles  1  and  7  are  alternate  exterior.  Name  one  other 
pair  of  alternate  exterior  angles.  Are  they  unequal  or  equal? 
Why? 

Use  a  sheet  of  thin  tracing  paper  and  carefully  copy 
your  drawing  of  Fig.  6,  using  thumb  tacks  to  secure  the 
work  in  position.  Move  the  copy  vertically  downward  by 
following    the    left-hand    margin.     Observe  the  direction 


FUNDAMENTAL  OPERATIONS  27 

relation  of  all  corresponding,  i.e.,  like  named  lines,  and 
the  magnitude  relations  of  all  correspondingly  numbered 
angles. 

Direction  axiom.  Two  lines  are  either  parallel,  perpen- 
dicular, or  oblique. 

Magnitude  axiom.  One  quantity  is  either  less  than,  equal 
to,  or  greater  than  another  quantity. 

Move  the  copy  horizontally  to  the  right  by  following 
the  lower  margin.  Observe  the  direction  of  all  corre- 
sponding lines  and  the  magnitude  relation  of  all  corre- 
spondingly numbered  angles. 

Restore  the  copy  or  tracing  to  its  original  position. 

Without  lifting  the  copy  move  it  through  90°,  keeping 
the  F  points  matched  by  pivoting  with  a  needle  point. 
The  AB  line  of  the  copy  will  then  be  at  right  angles  to  the 
AB  line  of  the  squared  paper.  Observe  the  direction  of  all 
corresponding  lines  and  the  relations  of  all  correspondingly 
numbered  angles. 

Ex.  40.  Locate  a  point  near  the  center  of  the  work  sheet 
and  mark  this  point  "  0."  Through  0  draw  a  horizontal  line 
H"  to  the  right.  Label  the  right  end  of  the  line  "A."  Set 
your  compass  to  extend  from  0  to  A  and  with  0  as  a  center  draw 
a  circle.  What  is  the  radius  of  the  circle  and  what  line  can  be 
used  to  represent  it.  Place  a  needle  point  at  0  and  the  straight- 
edge along  OA.  Rotate  the  straightedge  in  a  counter-clockwise 
direction  until  it  is  in  a  vertical  position,  and  represent  it  by  a 
line  OB.  What  kind  of  angle  and  how  many  degrees  have  been 
described  in  the  rotation?  OA  is  the  initial  side  of  the  angle, 
and  OB  is  the  terminal  side  sometimes  referred  to  as  the  radius 
vector  or  moving  arm.  An  acute  angle  is  greater  than  0°  but  less 
than  90°.  Continue  the  rotation  until  the  straightedge  is  hori- 
zontal and  draw  OC  the  terminal  side  of  the  straight  angle  thus 
formed.  CO  A  is  a  straight  angle  because  its  sides  OA  and  OC  are 
in  the  same  straight  line  with  the  vertex  0.  An  obtuse  angle 
is  greater  than  90°  but  less  than  180°. 

Referring  to  Fig.  7,  O  is  the  center  of  the  circle.  The 
perpendicular  diameters  CA  and  BF  divide  the  circle  into 


28 


PRACTICAL  MATHEMATICS 


four  quadrants  I,  II,  III,  and  IV,  and  also  indicate  the 
four  directions  which  are  designated  by  the  four  cardinal 
points  N,  E,  W,  S,  respectively.  If  we  stand  facing  the 
north  the  right  hand  will  extend  to  the  east. 

The  boundary  of  the  circle  is  called  the  circumference 
and  any  portion  of  the  latter,  such  as  LDA,  is  called  an 
arc,  i.e.,  an  arc  of  the  circle. 


Fig.  7. 


An  angle  such  as  LOA,  whose  vertex  is  at  the  center  of 
the  circle,  is  called  a  central  angle  and  its  sides  are  radii 
(plural  of  radius)  of  the  circle. 

A  radius  is  a  line,  such  as  OD,  joining  the  center  of  a 
circle  to  a  point  in  the  circumference.  A  chord  is  a  line 
joining  two  points  in  the  circumference.  A  diameter  equals 
twice  the  radius,  passes  through  the  center,  and  is  the 
longest  chord. 


FUNDAMENTAL  OPERATIONS  29 

An  inscribed  angle,  such  as  ACG,  has  its  center  on  the 
circumference  and  its  sides  are  chords  of  the  circle. 

An  arc  may  be  described  in  terms  of  the  angle  which 
intercepts  it  as  arc  LD  or  ^  Z  LOD  . 

The  circumference  of  the  circle  has  360°  of  arc,  and  each 
quadrant  contains  an  arc  of  90  arc  degrees  and  a  central 
angle  of  90  angle  degrees.  The  sum  of  the  angles  about 
the  central  point  0  equals  360°.  A  unit  central  angle 
intercepts,  i.e.,  cuts  off,  a  unit  arc.  Therefore  a  central 
angle  of  one  degree,  i.e.,  one-ninetieth  of  a  right  angle, 
intercepts  an  arc  of  one  degree,  i.e.,  one-ninetieth  of  a 
quadrant  arc.  A  central  angle  of  one  radian  (approxi- 
mately 57.3°)  intercepts  one  radian  of  arc  (a  length  equal 
to  the  radius).  A  central  angle  is  measured  by  the  arc 
which  it  intercepts. 

An  angle  may  be  measured,  therefore,  by  a  protractor 
by  placing  the  center  of  the  protractor  at  the  vertex  of  the 
angle  and  reading  the  number  of  degrees  intercepted  on 
the  arc. 

Ex.  41.  Referring  to  Fig.  7  designate  all  the  central  angles 
having  the  initial  side  OA,  and  state  also  their  respective  inter- 
cepted arcs.  These  angles  are  also  designated  as  angles  of  the 
I,  II,  III,  or  IV  quadrants,  if  their  terminal  sides  lie  in  these 
respective  quadrants.  Describe  the  angles  lying  in  each  quadrant 
of  Fig.  7. 

Ex.  42A.  What  will  be  the  position  of  the  terminal  side  of  the 
following  angles:  30°,  45°,  57.3°,  60°,  114.6°,  120°,  135°,  150°, 
210°,  240°,  270°,  300°,  360°? 

Ex.  42B.  Construct  the  angles  mentioned  in  Ex.  42A  and  with 
a  radius  of  2.25"  draw  a  concentric  circle,  i.e.,  a  circle  with  the 
same  center  0.  Extend  the  sides  of  the  angles  to  the  new  circum- 
ference. Does  this  extension  of  the  sides  of  the  angles  in  any 
way  effect  the  degree  measure  of  the  angles  or  the  measure  of 
their  respective  arcs  on  the  new  circumference? 

Ex.  43.  Designate  the  chords  in  Fig.  7.  Designate  the  inscribed 
angles  and  the  respective  arcs  which  they  intercept;  the  same  arcs 
are  also  intercepted  by  which  respective  central  angles?  Show 
by  measurement  that  each  of  the  inscribed  angles  is  measured 
by  one-half  of  its  respective  arc. 


30  PEACTICAL  MATHEMATICS 

A  line  such  as  FT  which  touches  a  circle  in  one  point  is  called 
a  tangent. 

Ex.  44.  What  is  the  greatest  value  which  the  inscribed  angle 
can  assume  when  CA  is  kept  fixed  and  the  chord  CG  is  rotated 
about  C?  Is  a  tangent  perpendicular  to  a  radius  drawn  to  its 
point  of  contact?    Why? 

Ex.  45.  Construct  a  tangent  to  a  circle  and  through  the 
point  of  contact  P  draw  a  chord  which  terminates  at  the  point 
R.  Determine  the  relation  of  the  angle  formed  by  tangent  and 
chord  compared  with  the  intercepted  arc. 

Ex.  46.  Draw  two  parallel  chords.  Measure  the  arcs  inter- 
cepted between  the  chords.     What  is  their  relation? 

Ex.  47.  Draw  a  tangent  and  a  parallel  chord.  What  is 
relation  between  all  the  arcs  intercepted? 

Ex.  48.  From  a  point  C  draw  two  tangents  to  a  circle  with 
center  0.  Measure  the  tangents,  the  angle  between  the  tangents 
and  the  central  angle  drawn  to  the  point  of  contact.  Draw  OC 
and  remeasure  the  central  angles.  By  what  name  would  you 
call  OC? 

Ex.  49.  From  a  point  E  draw  a  tangent  and  a  prolonged 
chord,  i.e.,  a  secant.  Measure  all  lines  and  arcs  and  state  their 
relation. 

Ex.  50.  Draw  two  angles  1  and  2  whose  sides  are  parallel 
and  extend  in  the  same  direction,  measure  them  and  state  their 
relation. 

Draw  an  angle  3  whose  sides  are  parallel  to  1  but  extending 
in  the  oppposite  directions,  measure  and  state  their  relation. 

Draw  an  angle  4  whose  sides  are  parallel  to  1  with  one  pair 
of  sides  extending  in  the  same  and  the  other  pair  in  opposite  direc- 
tions, measure  and  state  their  relation. 

Ex.  51.  Draw  two  acute  angles  5  and  6  with  their  sides 
perpendicular,  measure  and  state  their  relation. 

Draw  two  obtuse  angles  7  and  8  whose  sides  are  perpendicular, 
measure  and  state  their  relation. 

Draw  an  obtuse  angle  9  whose  sides  are  pendicular  to  5,  measure 
and  state  their  relation. 

Ex.  52.  The  Bisection  of  an  Angle.  On  the  sides  of  angle 
1  beginning  at  the  vertex  lay  off  a  unit  distance  (one  inch)  and 
at  these  points  draw  perpendiculars.  The  perpendiculars  (two 
straight  lines)  can  intersect  only  in  one  point.  The  line 
passing  through  the  intersection  and  the  vertex  is  the  bisector 
of  the  angle.  Draw  the  bisector  and  verify  by  measuring  the 
resulting   angles.     Measure   the   perpendiculars   and   state   their 


FUNDAMENTAL  OPEEATIONS  31 

relation.  From  the  unit  distant  points  as  centers  strike  arcs 
of  equal  radius.  Their  intersection  lies  on  the  bisector  of  the 
angle. 

Ex.  53.  The  Bisection  of  a  Line.  Draw  a  straight  line. 
Label  its  extremities  A  and  B,  respectively.  From  A  and  B 
strike  equal  arcs  which  intersect  above  and  below  the  line  at 
points  C  and  D  respectively.  A  line  passing  through  C  and  D  is 
the  perpendicular  bisector  of  AB. 

Ex.  54.  The  method  of  Ex.  53  may  be  used  to  draw  a  per- 
pendicular to  a  line  through  a  given  point  on  or  off  the  line,  AB. 
Select  a  point  C  off  the  line.  From  C  draw  any  arc  intersecting 
the  line  AB  at  E  and  F.  Use  E  and  F  as  centers  and  strike 
equal  arcs  intersecting  above  and  below  the  line  at  points  G  and 
H  respectively.  The  line  GH  is  the  perpendicular  to  the  given 
line  through  C.  Repeat  changing  the  position  of  C  so  that  it 
is  on  the  given  line. 

Ex.  55.  Draw  a  line  XY  about  4"  in  length.  Locate  a  point 
P  about  2"  above  the  middle  of  XY.  Draw  five  lines  from  P  to 
XY  two  of  which  shall  be  equal  and  one  of  which  shall  be  a  per- 
pendicular. Which  is  shortest?  Measure  and  record  any  other 
facts  of  interest  which  you  observe. 

8.  Axioms  and  Their  Applications.  Through  observa- 
tion, investigation,  and  reflection  we  learn  that  there  are 
many  fundamental  truths  which  are  axiomatic.  These  are 
formulated  in  groups  according  to  their  generality. 

Several  axioms  were  cited  above,  and  through  additional 
observation,  investigation,  and  reflection,  we  are  able  to 
appreciate  the  list  of  axioms  on  page  32. 

These  fundamental  truths  may  be  summarized  under  a 
single  statement  called  the  axiom  of  operations. 

Axiom  of  operations  (Ax.  Op.).  An  equality  is  pre- 
served when  a  like  operation  is  performed  upon  both 
members  of  an  equation,  otherwise  an  inequality  results. 

The  Ax.  Op.  includes  addition,  subtraction,  multiplica- 
tion, division,  power,  root,  and  trigonometric,  logarithmic, 
limiting,  derivative,  and  integral  operations.  By  their 
application  we  are  enabled  to  solve  equations,  i.e.,  deter- 
mine the  values  of  the    quantities    in   the    equations    as 


32 


PRACTICAL  MATHEMATICS 


Ss     +   O     O 


o    to     -©    h 

H 
3 


to 

II      II     II     -i     II  II 

M    M    «      |     «  Si 

+          e  ° 


CO  © 

+      "     nn 

CO  ,o    oo 

I    I    » 

r-^  ,0      ** 
II         H 

co  e 
I 
H 


§    ,©|-e    ^lio    3>  >    <£>    « 


el 


5s 

II 

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U8    hco 
>     oo 


03 

p 

cr 

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p  cr 

s8 


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CD  02 

I  i 

03  02 

O  03 


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cr:g 

cd  ^ 
0) 


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fa 

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FUNDAMENTAL  OPERATIONS 


33 


shown  below,  by  separating  a  specified  letter  from  all  the 
others. 


Given 

Result    Axiom  which  was 

Applied     Operator 

P  =  Zp 

P 
P  =  3 

Division 

Divisor  3 

P  =  Smp 

P 

1 1 

"      3m 

3mp  =  P 

P 

m  =  — 
3p 

1 1 

11      Sp 

P  =  EI 

,-J 

1 1 

"      I 

n 

*i 

( i 

11      E 

r  =  T-R 

T  =  r+R 

Addition 

Adjunct  R 

iJ- 

V  =  Ir    . 

Multiplication        Multiplier  r 

Ex.  56.  Prepare  a  list  of  the  equations  given  below;  show 
the  solution  for  each  letter,  stating  the  axiom  which  was  applied 
and  the  operator  used. 

E 

(a)  I  =—,  solve  for  E ,  and  from  this  result  solve  for  R. 

R 

(b)  V  =rl,  solve  for  I,  and  also  for  r. 

(c)  R  +r  =  T,  solve  for  r,  also  for  R. 

(d)  I  =EG,  solve  for  E,  also  for  G. 

vb 

(e)  5=—,  solve  for  rb,  and  from  the  result  solve  for  a,  r 


and  b. 


rb 


(/)        G=—,  proceed  as  in  (e)  and  solve  for  a,  r,  and  b. 

rb 
(g)         x=—,  proceed  as  in  (e)  and  solve  for  a,  r,  and  b. 


34  PRACTICAL  MATHEMATICS 

(h)         x  = — ,  proceed  as  in  (e)  and  solve  for  a,  b  and  R. 
a 

rl 
(i)         R  =  — ,  solve  for  A,  r  and  I J 

(j)         c  =  — ,  solv3  for  r. 
r 

Re 

(k)        E  =— ,  solve  for  r,  #,  and  e. 
r 

(I)         C  =—,  solve  for  c,  d,  D,  — ,  — . 

CL  C     U 

A  line  drawn  through  a  symbol  negatives  that  symbol, 
thus  a^b  means  a  is  unequal  to  b.  If  we  wish  to  give  the 
sense  of  the  inequality,  the  symbol  >  or  <  must  be  written 
between  the  quantities  with  the  larger  quantity  in  the 
opening  of  the  wedge  symbol.  a>b  means  a  is  greater 
than  b  and  x<  V  means  x  is  less  than  *f-. 

An  inequality  is  obtained  from  an  equality  when  any  of 
the  above  axioms  is  violated.  Suppose  we  have  25a  =  25a 
and  subtract  3a  from  one  side  and  6a  from  the  other.  Then 
these  two  subtractions  have  like  minuends  but  unlike 
subtrahends,  and  the  greater  remainder  results  from  the 
subtraction  of  the  lesser  subtrahend. 

Equality  25a  =  25a, 
lesser  subtrahend  3a<6a  greater  subtrahend, 
greater  remainder  22a  >  19a  lesser  remainder. 

Ex.  57.  Perform  the  operations  of  addition,  multiplication 
division,  power  and  root  with  unequal  operators  on  the  equation 
25a  =  25a.  State  the  relation  of  the  results.  Can  the  Axiom  of 
Operations  be  extended  to  inequalities  providing  the  operators 
are  equal? 

In  addition  to  the  axioms  of  operation  we  have  those 
observed  from  the  construction  work  as  follows : 


FUNDAMENTAL  OPERATIONS  35 

Magnitude  Axiom.  One  quantity  is  either  less  than, 
equal  to  or  greater  than  another  quantity.  The  whole  is 
equal  to  the  sum  of  its  parts.  A  quantity  may  be  sub- 
stituted for  its  equal  in  any  equation  or  inequality. 

Direction  Axiom.  Two  lines  are  either  parallel, 
perpendicular,  or  oblique.  Any  line  has  the  same  direction 
as  a  line  with  which  it  coincides. 

Point  line  Axiom.  A  point  is  or  is  not  on  a  line. 
Between  two  points  only  one  straight  line  can  be 
drawn.  Two  straight  lines  intersect  in  a  point.  Through 
a  point  only  one  line  can  be  drawn  parallel  to  a  given 
line. 

Construction  Axiom.  Any  construction  line  may  be 
added  to  a  geometric  figure  provided  its  relation  does 
not  contradict  or  violate  the  relation  of  the  parts  of  the 
given  figure. 

Motion  Axiom.  An  entire  figure  or  part  thereof  may 
be  moved  about  in  a  plane  or  in  space  and  restored  or 
superposed  in  any  other  position  provided  the  essential 
relations  of  the  parts  are  unaltered. 

Plane  Axiom.  A  plane  is  determined  by  three  points 
not  in  the  same  straight  line;  by  a  line  and  a  point 
outside  of  it;  by  two  intersecting  lines;  by  two  parallel 
lines. 

Reductio  ad  absurdum  (Red.  ad  ab.)  Axiom.  A 
statement  of  relationship  is  validated  when  all  contradictory 
or  opposite  constructions  have  been  falsified,  i.e.,  reduced 
to  an  absurdity. 

9.  Summary  of  Theorems  on  Lines  and  Angles.  A 
theorem  is  unlike  an  axiom  because  it  is  a  truth  admitting 
of  or  requiring  demonstration.  A  demonstration  is  a 
formal  array  of  statements  each  in  logical  sequence  and  each 
validated  by  the  sufficient  necessary  authorities.  The  list 
of  proper  mathematic  authorities  to  substantiate  mathe- 
matic  statements  includes: 

First,  the  given  stated  condition  called  the  hypothesis. 


36  PRACTICAL  MATHEMATICS 

Second,  axioms  of  equations,  inequalities,  constructions, 
relation,  motion,  and  position. 

Third,  accurately  formulated  definitions. 

Fourth,  a  theorem  previously  demonstrated. 

Geometry  is  one  of  the  most  beautiful  structures  of 
thought  in  which  the  above  cited  authorities  secure  the 
safety  of  the  mind's  edifice.  Although  the  theorems  may 
be  discovered  in  any  order,  it  is  essential  that  they  should 
be  demonstrated  in  a  systematic  manner,  owing  to  their 
interdependence.  By  definition  a  straight  angle,  as  its 
name  implies,  is  an  angle  whose  sides  are  in  the  same  straight 
line  with  the  vertex.  In  Fig.  6,  AEB  is  a  straight  angle, 
because  its  sides  EA  and  EB  are  in  the  same  straight  line 
with  the  vertex  E.  By  measurement  we  discover  that 
AEB=\m°.  Do  all  straight  angles  =  180°?  The  theorem 
establishes  the  truth  when  demonstrated  that  every  straight 
angle  equals  180°. 

The  following  theorems  are  formulated  from  the  observa- 
tions made  above.  They  are  so  simple,  in  fact,  that  they 
are  often  accepted  without  demonstration.  They  are  given 
in  the  order  in  which  they  would  have  to  be  demonstrated. 
If  the  student  has  not  recognized  them,  he  should  review 
the  preceding  work  until  he  is  entirely  familiar  with  their 
significance,  by  experiment,  construction,  and  measurement. 

1.  A  straight  angle  equals  180°. 

2.  Adjacent    angles    whose    exterior    sides    are    in    a 
straight  line  are  supplementary. 

3.  Supplementary  adjacent  angles  have  their  exterior 

sides  in  a  straight  line. 

4.  Vertical  angles  are  equal. 

5.  At  a  point  in  a  line  only  one  perpendicular  can  be 

drawn  through  the  line. 

6.  From  a  point  outside  a  line  only  one  perpendicular 

can  be  drawn  to  the  line. 

7.  Any  point  on  a  perpendicular  erected  at  the  middle 


FUNDAMENTAL  OPERATIONS  37 

of  a  line  is  equally  distant  from  the  extremities 
of  the  line. 

8.  Any  point  not  on  a  perpendicular  erected  at  the 
middle  of  a  line  is  unequally  distant  from  the 
extremities  of  the  line. 

9.  A  perpendicular  is  the  shortest  line  which  can  be 

drawn  from  a  point  to  a  line. 

10.  Lines  perpendicular  to  the  same  line  are  parallel. 

11.  If  one  of  a  number  of  parallel  lines  is  perpendicular 
to  a  line,  the  others  are  perpendicular  to  the  same 
line. 

12.  Lines  parallel  to  the  same  line  are  parallel  to  each 
other. 

When  If 

two     parallel     lines    14.  Alternate  -  interior   angles 
are  cut  by  a  trans-  are  equal,  or 

versal,  then 

13.  Alternate-interior  16.  Exterior-interior  angles  are 
angles  are  equal  equal,  or 

15.  Exterior-interior  18.  Unilateral-interior    angles 

angles  are  equal  are  supplementary,  or 

17.  Unilateral -interior  20.  Alternate-exterior  angles 
angles   are   supple-  are  equal, 

mentary  then  the  two  lines  cut  by  the 

19.  Alternate-exterior  transversal  are    parallel. 

angles  are  equal 

21.  Angles  are  equal  when  their  pairs  of  sides  are  parallel 

and  extend  in  the  same  or  opposite  direction. 
They  are  supplementary  when  one  pair  of  parallel 
sides  extend  in  the  opposite  direction  from  the 
vertex. 

22.  Angles  are  equal  when  their  pairs  of  sides  are  per- 
pendicular and  both  are  acute  or  both  obtuse. 
They  are  supplementary  when  one  of  these  angles 
is  acute  and  the  other  obtuse. 


38 


PRACTICAL  MATHEMATICS 


23.  Every  point  in  the  bisector  of  an  angle  is  equally 
distant  from  the  sides  of  the  angle. 
10.  Demonstration.      The  demonstration  of  theorem  1 
which  follows  illustrates  the  rigorous  formal  proof  and  is 
typical  of  all  logical  methods  of  mental  discipline. 


I  Theorem  1 

(Theorem)  A   STRAIGHT  ANGLE   EQUALS  180° 

X 


(hypothesis) 
(conclusion) 


Given  St  Z  A  CD 
Prove  ZACD  =  180° 
(1)  ACD  is  a  St  Z. _ 


(2)  /.     AD  is  a  St  line. 

(3)  Draw  XC±AD 

(proof)         (4)  Then  ZACX  =  90c 


(5)  andZXCD     =90° 

(6)  /.      ZACX+ZXCD  =  180°. 


(7)  but 

(8)  /. 


ZACD  =  ZACX+  ZXCD 
ZACD  =  180° 


(authorities) 

-Hyp 
_DefStZ 
Cons  Ax 


Def  J_ 

-Add  Ax 
_Mag  Ax 
=ty  Ax 


Therefore  a  straight  angle  equals  180°. 

Explanation.  The  Arabic  numeral  1  in  the  heading 
stands  for  the  first  theorem  in  Book  I.  A  book  was  the 
Greek  equivalent  of  our  modern  designation  of  a  chapter 
or  part,  and  in  geometry  has  been  tenaciously  adhered  to 
since  the  time  of  Euclid. 

The  theorem  states  the  facts  as  they  have  been  observed, 
developed,  and  formulated  in  the  constructions  and  measure- 
ments. 

The  demonstration  proves  the  fact  generally,  i.e.,  for 
all  straight  angles. 


FUNDAMENTAL  OPERATIONS  39 

The  figure  is  a  linear  representation  of  the  thing  men- 
tioned in  the  hypothesis,  and  in  this  case  the  picture  of  the 
solid  lined  straight  angle  ACD  is  a  mental  aid  in  keeping 
the  condition  before  us. 

A  theorem  is  always  analyzed  into  two  parts:  hypothesis 
and  conclusion.  These  are  stated  immediately  after  the 
figure. 

The  hypothesis  begins  with  the  words  given,  and  there- 
fore relates  the  known  things  both  by  definition  name  and 
by  figure  name. 

The  conclusion  is  the  fact  which  is  to  be  established. 

The  numbered  statements  begin  with  a  recitation  of 
the  fact  stated  in  the  hypothesis  (Hyp),  and  are  validated 
by  the  abbreviated  authorities  which  are  placed  at  the 
right  end  of  their  respective  lines. 

(1)  states  that  /.ACD  is  a  straight  Z.  Upon  investi- 
gation of  the  definition  of  a  straight  angle  (St  Z)  we  are 
led  to  make  statement  (2),  which  is  authorized  by  quoting 
the  definition  of  a  straight  angle  (Def  St  Z). 

We  would  be  at  a  standstill  now  if  it  were  not  for  the 
fact  that  Ave  have  learned  that  a  right  angle  has  90°.  Bub 
twice  90°  equals  180°  and  this  seems  to  suggest  building 
two  right  angles  out  of  /.ACD. 

In  (3)  we  draw  the  dash  (construction)  line  XC  perpen- 
dicular (J_)  to  AD  at  C.  We  are  permitted  to  do  this 
bjr  the  Construction  Axiom  (Cons  Ax)  which  states  that 
the  solid  lines  may  be  supplemented  by  any  construction 
lines  (dash)  which  do  not  represent  conditions  contrary  to 
the  hypothesis.  This  construction  creates  /s  ACX  and 
XCD. 

(4)  and  (5)  state  the  degree  measure  of  Zs  ACX  and 
XCD  from  the  definition  of  a  perpendicular  (Def  J_ ) . 

In  (6)  the  sum  of  Zs  ACX  and  XCD  equals  the  sum  of 
their  equals.  Therefore  /ACX+  / XCD  =  90°+90°  =  180°. 
We  have  quoted  the  Addition  Axiom  which  says  if  equals 
are  added  to  equals  their  sums  are  equal. 


40  PRACTICAL  MATHEMATICS 

(7)  brings  in  an  independent  fact  suggested  by  the 
examination  of  the  figure  in  which  it  is  evident  that  the 
whole  angle  ACD  is  composed  of  the  two  angles  ACX  and 
XCD.  This  statement  is  therefore  backed  by  the  Mag- 
nitude Axiom  (Mag  Ax)  which  states  that  the  whole  equals 
the  sum  of  its  parts. 

The  Equality  Axiom  states  that  things  equal  to  the  same 
thing  are  equal  to  each  other  (  =  ty  Ax).  Taking  the  state- 
ments (6)  and  (7)  together  we  form  the  new  equality  (8). 

This  ends  the  proof  because  we  have  arrived  at  the 
conclusion. 

Having  conclusively  established  the  theorem  it  is  con- 
sistent and  even  preferable  to  write  it  at  the  end  beginning 
with  the  word  therefore. 

11.  Triangles,  Quadrilaterals  and  Polygdns.  In  general, 
if  we  have  two  or  more  figures  of  the  same  kind,  and  can 
make  them  coincide,  i.e.,  place  one  over  the  other  into 
exact  correspondence  in  position,  they  are  said  to  be  equal 
figures.  In  each  figure  the  corresponding  lines  will  be  equal 
and  the  corresponding  angles  will  be  equal.  This  prin- 
ciple enables  us  to  measure  a  line  or  lay  off  a  distance 
with  the  dividers.  The  points  of  the  dividers  are  adjusted 
to  the  extremities  of  the  line  and  without  alteration  they 
are  applied  to  a  scale.  The  measure  of  the  distance  between 
divider  points  is  also  the  measure  of  the  length  of  the 
line.     This  is  illustrated  in  Fig.  8. 

It  would  be  exceedingly  laborious  were  we  compelled  to 
resort  to  this  matching  process  whenever  two  triangles 
were  to  be  proven  equal.  There  are  six  parts  to  every 
triangle,  viz:  the  three  angles  and  the  three  sides.  If 
three  of  these  parts,  including  at  least  one  side,  are  respect- 
ively equal,  then  the  triangles  are  equal. 

Ex.  58.    Make  a  list  of  all  the  lettered  triangles  in  Fig.  8. 

Ex.  59.  Draw  an  acute  angle  1,  with  vertex  0,  and  unequal 
sides  AO  and  OB.  Measure  1,  AO,  OB.  On  a  sheet  of  tracing 
paper  draw  a  line  ED  =  OB  and  parallel  to  it.     Through  E  draw 


FUNDAMENTAL  OPERATIONS  41 

EF  parallel  to  and  =  to  AO,  forming  angle  1'.  Why  is  1=1'? 
Why  are  the  sides  equal  in  length?  Join  A  with  B  forming  triangle 
AOB  and  join  D  with  F  forming  triangle  FED.  Now  place  AOB 
upon  DEF  so  that  the  equal  parts  coincide  or  match.  Do 
all  six  parts  of  the  two  triangles  coincide?  Therefore  the  two 
triangles  are  in  what  relation? 

Construct  a  triangle  GHJ,  measure  GH,  and  the  angles  GHJ 
and  HGJ.  On  tracing  paper  construct  a  triangle  KLM,  making 
KL  =GH,  and  angles  KLM  =GHJ,  and  LKM  =HGJ,  respectively. 
Two  angles  may  be  laid  off  equal  by  making  their  sides  respect- 
ively parallel  or  perpendicular,  according  to  theorems  21  and  22, 
or  using  the  protractor.  We  may  construct  two  angles  equal 
by  laying  off  equal  arcs  of  equal  radius,  using  the  vertices  as 
centers.  See  Fig.  9  /  and  /'.  Will  these  triangles  coincide? 
Therefore  the  two  triangles  are  in  what  relation? 

Construct  a  triangle  PQR,  and  measure  its  three  sides  PQ, 
QR,  and  RP.  On  tracing  paper  draw  ST  =PQ,  from  S  as  a  center 
strike  an  arc  whose  radius  equals  PR;  from  T  as  a  center  strike 
an  arc  whose  radius  equals  QR.  At  the  intersection  of  the  two 
arcs  place  the  letter  V.  Join  V  with  S  and  T,  forming  triangle 
STV.  Will  triangle  STV  coincide  with  PQR  t  Therefore  the  two 
triangles  are  in  what  relation? 

Two  triangles  are  equal  when  they  have  the  following 
parts  respectively  equal  as  illustrated  in  Fig.  9 : 

One  side  and  two  angles  /  /', 

Two  sides  and  one  angle,  II  II', 

Three  sides,  III  IIP. 

In  a  right-angled  triangle  this  reduces  to, 

One  side  and  an  acute  angle,  IV  IV, 

A  leg  and  the  hypotenuse,  V  V, 

Two  legs,  VI  VI'. 

A  right  triangle  is  one  having  a  right  angle.  The 
side  opposite  the  right  angle  is  called  the  hypotenuse 
and  the  other  two  sides  are  called  legs  or  arms  or  simply 
sides. 

Ex.  60.  Draw  two  parallel  lines  UW  and  XY.  On  UW 
locate  a  point  Z,  and  connect  Z  with  X  and  F.  Show  that 
UZX=ZXY,  also  WZY  =ZYX. 


42 


PEACTICAL  MATHEMATICS 


B' 

C  C 

Fig.  9. — Equal  Triangles. 


FUNDAMENTAL  OPERATIONS  43 

See  theorem  13.  But  Zs  UZX+XZY+WZY  =  180°. 
Why?  Replace  the  Zs  UZX  and  WZY  by  their  equal 
angles  in  the  triangles.     Therefore: 

The  sum  of  the  angles  of  a  triangle  equals  180°. 

Ex.  61.  In  the  preceding  figure  extend  XY  to  a  point  V. 
An  exterior  angle  ZYV  of  a  triangle  is  formed  between  a  side 
and  an  adjacent  side  produced  or  extended.  Show  thatZsZFF 
=ZXY  +ZXY.    Therefore : 

The  exterior  angle  of  a  triangle  is  equal  to  the  sum  of  the  two 
non-adjacent  interior  angles. 

A  four-sided  figure  is  called  a  quadrilateral. 

A  quadrilateral  whose  opposite  sides  are  parallel  is 
called  a  parallelogram.  Construct  a  parallelogram  and 
measure  its  opposite  sides.  What  is  their  relation?  .  The 
diagonals  are  constructed  by  joining  the  opposite  vertices. 
The  intersection  of  the  diagonals  divides  them  in  what 
relation?  One  diagonal  will  divide  the  parallelogram  into 
two  equal  triangles.  Why?  A  trapezoid  is  a  quadrilateral 
with  one  pair  of  parallel  sides. 

Ex.  62.  Specify  all  the  lettered  quadrilaterals  in  Fig.  8  both 
by  title  and  figure  name. 

Ex.  63.  In  Fig.  8,  the  dash  lines  are  drawn  parallel  through 
the  equidistant  points  1,  1,  1  ...  1  on  MN.  Every  trans- 
versal of  these  parallels  will  be  subdivided  (multisected)  into  uni- 
formly spaced  divisions.  If  two  transversals  are  parallel  their 
sections  called  segments  are  mutually  equal.  Check  the  point 
spacing  on  all  transversals  by  means  of  bow  dividers  as  shown 
in  the  figure. 

Ex.  64.  In  Fig.  8  how  many  paralellograms  are  formed  by 
the  dash  lines? 

Ex.  65.  In  Fig.  8  how  many  trapezoids  are  formed  by  the 
dash  lines? 

Ex.  66.  In  Fig.  8  how  many  triangles  are  formed  by  the 
dash  lines? 

Triangles  are  similar,  i.e.,  of  the  same  shape  under  any  one 
of  the  following  conditions: 

When  their  homologous  (like  placed)  angles  are  equal, 

When  their  homologous  sides  are  parallel, 


44  PRACTICAL  MATHEMATICS 

When  the  homologous  sides  are  perpendicular, 

When  their  homologous  sides  are  in  the  same  ratio,  i.e., 

proportional. 
Ex.  67.    In  Fig.  8  there  are  how  many  groups  of  similar  triangles 
and  how  many  similar  triangles  in  each  group? 

A  many  sided  figure  is  called  a  polygon  (n-gon).  A 
limited  number  of  polygons  have  special  names  according 
to  the  following  list: 

Number  of  Sides 

or  Angles  Name  Figure-gon 

3  triangle,  trilateral,  or  trigon  3-gon 

4  quadrilateral,  quadrigon  4-gon 

5  pentagon  5-gon 

6  hexagon  6-gon 

7  heptagon  7-gon 

8  octagon  8-gon 

9  nonagon  9-gon 

10  decagon  10-gon 

11  j  undecagon  11-gon 

12  duodecagon  12-gon 
15  pentadecagon  1 5-gon 
20              icosagon                                            20-gon 

Regular  polygons  have  equal  sides  and  equal  angles. 
Regular  polygons  are  inscribed  (constructed  internally)  in  a 
circle  by  laying  off  equal  chords,  striking  equal  arcs,  or 
forming  equal  central  angles. 

An  irregular  polygon  may  result  from  an  inequality  of 
sides,  or  an  inequality  of  angles,  or  by  the  substitution 
of  arcs  for  any  of  its  sides. 

Ex.  68.  Name  the  pairs  of  regular  and  irregular  polygons 
in  Fig.  10. 

Polygons  are  similar  when  their  corresponding  sides 
are  proportional  and  their  corresponding  angles  are  equal. 
Regular  polygons  of  the  same  name  are  similar.  A  set 
of  similar  figures  can  be  decomposed  into  another  set  of 


FUNDAMENTAL  OPERATIONS 


45 


-   ''                 .s' 

•^ 

i^~  /  ' 

iK^fi  ?/     "\     * 

x  'S^ 

/ 

46  PRACTICAL  MATHEMATICS 

similar  figures  by  joining  corresponding  points  in  the  first 
set. 

Ex.  69.  Name  the  similar  polygons  in  Fig.  10?  Are  poly- 
gons similar  when  their  sides  are  proportional  and  also  either 
parallel  or  perpendicular? 

Ex.  70.    Why  are  the  triangles  in  Fig.  10  similar? 

Ex.  71.  Which  of  the  polygons  in  Fig.  10  show  an  exterior 
angle,  what  is  its  relation  to  the  central  angle  subtended  by  the 
side  which  was  prolonged? 

Ex.  72.  What  is  the  value  of  the  central  angle  for  constructing 
the  chord  of  each  regular  polygon  in  Fig.  10?     * 

Ex.  73.  What  is  the  value  of  the  angle  formed  by  adjacent 
sides  in  each  polygon  of  Fig.  10? 

Ex.  74.  Suppose  we  designate  the  number  of  sides  of  the  poly- 
gon by  n,  then  show  that  the  sum  of  the  angles  of  the  polygons 
in  each  case  will  be  (n—  2)  times  180°.  Verify  by  constructing 
enlarged  figures  and  measuring.  Also  do  the  same  for  irregular 
figures. 

Ex.  75.  Construct  the  exterior  angles  to  the  figures  in  Ex. 
74  and  show  that  the  sum  of  the  exterior  angles  of  a  polygon 
equals  two  straight  angles. 

Ex.  76.  Construct  a  parallelogram.  Measure  two  non- 
parallel  sides  and  the  angle  formed,  i.e.,  included  by  these  two 
sides.  With  these  measurements  construct  an  equal  parallelo- 
gram.   Therefore : 

Two  parallelograms  are  equal  when  they  have  two  sides  and 
the  included  angle  respectively  equal. 

When  the  sides  of  a  parallelogram  are  rectified,  i.e., 
drawn  perpendicular,  then  the  figure  is  a  rectangle. 

When  the  sides  of  a  rectangle  are  made  equal,  then 
the  figure  is  called  a  square. 

12.  Areas  of  Figures. 

Ex.  77.  On  a  sheet  of  squared  paper  ink  in  one  of  the  smallest 
squares.  If  we  consider  its  sides  to  be  one  unit  of  length  then 
the  surface  bounded  by  the  four  unit  lines  is  called  a  unit  of 
area.  Construct  four  other  squares  whose  respective  sides  are 
two  units,  three  units,  five  units,  ten  units.  How  many  unit 
squares,  i.e.,  units  of  area  are  there  in  each  of  the  larger  squares? 
Designate  these  figures  with  Roman  notation  calling    the    unit 


FUNDAMENTAL  OPERATIONS  47 

square  I,  and  the  other  squares  II,  III,  IV,  and  V  respectively. 
What  is  the  relation  of  the  magnitudes  obtained  by  measuring 
the  area,  and  the  magnitude  obtained  by  measuring  the  sides 
of  each  square?  If  the  side  of  a  square  measures  (contains)  a 
units,  how  many  unit  areas  are  there  in  each  row  and  how  many 
unit  areas  in  each  column  or  file?  Irrespective  of  how  we  count 
the  unit  squares,  we  shall  find  there  are  a  times  a  of  them,  or  a2 
unit  areas.  It  is  originally  due  to  this  fact  that  the  exponent 
2  over  a  letter  or  a  number  is  read  as  the  square  of  that  letter 
or  number.     Therefore: 

The  area  of  a  square  measured  in  square  units  is  the  square  of 
its  side  measured  in  linear  units. 

The  linear  unit,  i.e.,  unit  of  length,  may  be  a  foot, 
an  inch,  a  centimeter,  or  any  other  arbitrary  length  which 
we  may  choose  for  convenience.  A  centimeter  is  approx- 
imately equal  to  2.54  ins.  and  is  one  one-hundredth  of  a 
meter.  Show  that  a  meter  is  approximately  equal  to  39.37 
ins. 

Consider  the  side  of  a  square  V  is  one  unit  in  length, 
then  the  area  of  square  V  is  one  unit.  Why?  The  area 
of  I  then  becomes  a  subunit  and  is  numerically  equal  to 
one  one-hundredth  of  the  unit  area  V. 

Ex.  78.  Express  the  area  of  squares  II,  III,  IV,  in  terms  of 
the  unit  area  V. 

Ex.  79.  Consider  square  II  as  the  unit  of  area  and  express 
the  area  of  IV  in  terms  of  II. 

We  may  define  the  area  of  a  figure  as  the  ratio  of  that 
figure  to  the  unit  figure.  A  ratio  is  written  as  a  fraction; 
the  numerator  expresses  the  magnitude  of  the  first  mentioned 
figure  or  quantity,  and  the  denominator  expresses  the 
magnitude  of  the  second  mentioned  figure  or  quantity. 
By  the  definition  of  area  we  have : 

„  TTT     IV    magnitude  of  IV    V- 
Area  of  IV  =  — •=      &  .      _ — -—  =  -f  =  6.25  units. 
II     magnitude  of  II       I 

The  magnitude  of  a  unit  is  one. 


48  PRACTICAL  MATHEMATICS 

Ex.  80.  On  squared  paper  construct  a  rectangle  3.5  ins.  long 
and  1.5  ins.  wide.  Select  any  convenient  unit  for  measuring  the 
sides,  and  the  corresponding  unit  for  measuring  the  area.  Repeat 
using  a  different  linear  unit  with  a  corresponding  square  unit. 
By  means  of  the  ruled  lines  on  the  paper  measure  the  sides  and 
area  in  all  possible  ways.  Write  the  law  for  expressing  the  area 
of  the  rectangle  in  terms  of  its  length  and  width  (breadth). 
Designate  the  length  by  1,  the  width  by  w  and  the  area  by  A. 
When  the  law  is  abbreviated  by  using  letters  and  symbols  it 
becomes  a  formula.  Show  why  A  =lw  is  the  formula  for  obtaining 
the  area  of  a  rectangle.  Determine  the  area  of  the  above  rectangle 
by  substituting  the  values  of  1  and  w  in  inch  units?  A  =3.5  X  1.5 
=5.25  sq.in. 

Ex.  81.  In  the  Ex.  80,  if  either  I  or  w  is  in  error  by  1  per 
cent,  what  is  the  error  in  A?  Per  cent  is  a  contraction  of  the 
Latin  per  centum,  which  means  by  the  hundred.  One  per  cent 
means  one  one-hundredth  or  one  part  in  one  hundred,  and  is 
usually  written  1%.  Therefore  1%  of  3.5  means  3.5  times  1/100  or 
3.5/100,  or  0.035,  and  is  the  amount  by  which  I  is  too  large  or  too 
small. 

Ex.  82.  In  the  Ex.  80,  if  both  I  and  w  are  1%  in  error,  what 
is  the  per  cent  error  in  A? 

Observation.  The  product  of  several  factors  has  as  many 
dependable  figures  as  the  least  number  of  dependable  figures 
in  any  of  its  factors. 

Ex.  83.  Near  the  left  end  of  a  horizontal  line  of  the  squared 
paper  locate  a  point  A  and  3  ins.  to  the  right  of  A  locate  B. 
One  inch  vertically  above  A  and  B  respectively,  locate  D  and  C. 
What  kind  of  figure  is  ABCD?  How  many  unit  squares  does  it 
contain?  What  is  its  area  in  square  inches?  Through  A  draw 
a  line  to  the  right,  making  an  angle  of  60°  with  AB,  and  mark  E 
at  its  intersection  with  DC.  Through  B  draw  a  parallel  to  AE 
intersecting  DC  produced  in  the  point  F.  What  kind  of  a  figure 
is  ABFE?  The  quadrilateral  ABCE  is  a  trapezoid.  Which  pair 
of  opposite  sides  in  ABCE  is  parallel?  What  magnitude  relations 
exist  between  triangles  ADE  and  BCF?  Why?  By  means  of 
the  trapezoid  ABCE,  and  the  addition  of  the  one  or  the  other 
of  the  triangles  ADE  and  BCF,  show  that  the  parallelogram 
ABFE  equals  the  rectangle  ABCD.  Check  by  counting  the  unit 
squares  and  their  fractional  parts  included  in  ABFE. 


FUNDAMENTAL  OPERATIONS  49 

The  dimensions  of  A  BCD  are  its  length  and  width, 
which  are  the  measure  of  its  two  perpendicular  sides  AB 
and  AD.  In  the  case  of  the  parallelogram  ABFE,  we 
can  measure  the  side  AB  for  its  length,  but  its  width,  also 
called  altitude,  is  measured  on  a  line  perpendicular  to 
AB  and  DC.  The  base  is  the  side  on  which  the  figure 
rests.     Therefore : 

The  area  of  a  parallelogram  is  equal  to  the  product 
of  its  length  times  its  width  or  the  product  of  its 
base  times  its  altitude. 

Ex.  84.  Supplement  the  construction  of  Ex.  83  by  connecting 
A  with  F,  dividing  ABFE  into  equal  triangles  ABF  and  AEF. 
Why?     Therefore: 

The   area   of   a   triangle   equals   one-half  the   product  of  its 
base  times  its  altitude. 

The  base  of  a  triangle  is  any  one  of  its  sides,  and  cor- 
respondingly its  altitude  is  its  perpendicular  distance  from 
this  side  to  the  opposite  vertex. 

Since  ABFD  is  not  a  parallelogram,  the  diagonal 
AF  does  not  bisect  it.  The  area  of  ABFD  is  the  sum  of 
the  areas  of  triangles  ABF  and  ADF.  Designate  the  bases 
AB  and  DF  by  a  and  b  respectively,  and  the  altitude  by 
h.  What  two  lines  in  the  figure  can  be  used  to  measure 
the  altitude  of  both  triangles?  Why?  The  area  of  tri- 
angles ABF  and  ADF  are  i  ah  and  i  bh,  respectively. 
Therefore  the  area  of  AJBFD  =  *ah+ibh  =  ih(a+b);  from 
this  formula  we  interpret : 

The  area  of  a  trapezoid  equals  one-half  the  product  of 
its  altitude  times  the  sum  of  the  bases  or  parallel  sides. 

The  altitude  of  the  trapezoid  is  the  perpendicular  dis- 
tance between  its  parallel  sides.     The  median  of  a  trapezoid 


50 


PRACTICAL  MATHEMATICS 


joins  the  mid-points  of  its  non-parallel  sides  and  is  parallel 
to  its  bases.     See  Fig.  11. 

Ex.  85.  Supplement  the  figure  of  Ex.  84  by  connecting  B 
with  D  and  also  with  E,  and  connect  A  with  C.  Join  G  and  H 
the  midpoints  of  DA  and  FB  respectively.  Rule  and  label  the 
following  "  TABLE  I.  LENGTHS."  Under  the  columns  headed 
lines,  write  the  figure  names  of  all  lines,  after  measuring  each 


Fig.  11. 

line  write    its    numeric  value  in    the  next   column,  alongside  of 
the  name  to  which  it  belongs.     Rule  the  table  with  a  straightedge. 


TABLE  I.     LENGTHS 


Line. 

Length, 
Inches. 

Line. 

Length, 
Inches. 

AB 
DF 
DE 
DC 
EC 
EF 
CF 
GH 

3.00 

AD 
AE 
AC 
AF 
BD 
BE 
BC 
BF 

1.00 

Rule  and  label  the  following  "  TABLE  II.  AREAS  OF  QUAD- 
RILATERALS." Under  the  heading  bases,  write  the  numeric 
values,  being  careful  to  note  that  the  trapezoid  has  two  bases 
which  are  recorded  with  a  comma  between  them.  Under  the 
columns  headed  altitude  and  area,  write  the  respective  values 
in  the  same  horizontal  line  with  the  figure  to  which  they  belong. 
Fill  in  other  blank  spaces  with  the  names  of  the  quadrilaterals. 


FUNDAMENTAL  OPEEATIONS 
TABLE  II.    AREAS  OF  QUADRILATERALS 


51 


Quadrilateral.' 

Figure. 

Bases. 

Altitude. 

Area. 

trapezoid 

DGHF 

<  < 

GABH 
ABCD 
ABFE 
ABED 
ABCE 
ADFB 
ABFC 

Rule  and  label  the  following  "TABLE  III.  AREAS  OF 
TRIANGLES,"  and  make  the  entries  in  the  respective  columns. 
The  headings  of  the  last  four  columns  are  a  duplication  of  those 
of  the  first  four  columns.  This  is  done  to  save  space  in  the  vertical 
direction  when  there  are  many  entries.  The  symbol  A  read 
"  delta,"  means  triangle;  base,  altitude,  and  area  are  abbreviated 
by  b,  h,  and  A,  respectively. 

TABLE  III.     AREAS  OF  TRIANGLES 


A 

b 

h 

A 

A 

b 

h 

A 

ADE 

ACB 

ADC 

AFB 

ADF 

BDE 

AEC 

BDC 

AEF 

BDF 

ACF 

BEF 

AEB 

BEC 
BCF 

13.  Variation,  Ratio,  and  Proportion.  In  Table  III  we 
observe  that  all  the  triangles  have  an  equal  altitude.  As 
a  check  on  the  work  of  multiplication,  show  that  for  all 
triangles  of  equal  altitude  the  area  divided  by  the  base 
equals  the  constant,  i.e.,  fixed  value  of  half  the  altitude. 
In  comparing  a  number  of  these  triangles,  what  words 
should  be  supplied  to  complete  the  following  statement: 
in   triangles   of   equal   altitude,   that   one   which   has  the 


52  PRACTICAL  MATHEMATICS 

greatest  base  has  the  .  .  .  area  and  correspondingly  that 
one  which  has  the  .  .  .  base  has  the  least  area.  This  is 
a  law  of  direct  variation,  and  may  be  symbolized  by  use 
of  the  variation  symbol  (oc)  written  between  the  area  and 
the  base.     Therefore  from  the  law  of  variation  we  write: 

A  oc  b,    but  A  =  \hb  =  constant  times  0. 

Observation.     The  variation  symbol  may  be  replaced  by  an 
equal  sign  and  a  constant. 

Again  referring  to  the  earlier  statement  above  we  write : 

area  of  a  triangle        A     h     altitude 


base  of  same  triangle     b     2 


Since  A  is  the  symbol  for  area  in  general,  a  necessity 
arises  for  distinguishing  the  A's  of  the  different  triangles. 
This  is  provided  for  by  attaching  a  small  number  called 
a  subscript  at  the  lower  right  side  of  the  letter  thus:  A\, 
A 2,  A3,  would  represent  the  respective  areas  of  triangles 
ADC,  ADE,  ADF.  In  corresponding  manner  the  bases  of 
these  triangles  would  be  designated  by  b\,  b2,  03,  and  the 
altitudes  by  hi,  h2,  ^3.  Therefore  we  write  like  subscripts 
for  all  parts  of  the  same  figure : 

A\    hi      A2_h2%    A%_hz. 
0j  ~~2  ;     b2~  2'     63  ~2 

but  since  hi=h2  =  h3,  then  by   =ty  Ax,   the  three  ratios 

A\    A2    As  .  A\    A2    A3     ™  . 

~    t^,   1—,  are  equal:    or  it-=t"s=1~-     Tms  1S  a  con~ 

Ol        02        03  01         02         03 

tinued  equality,  also  a  continued  proportion,  and  is  at  the 
same  time  a  condensed  form  of  three  distinct  equations, 
viz.: 


Ai    A2 

Ai    .43 

A2    A3 

61      62 ' 

01      03 ' 

02        03 

The  above  equations  are  proportions  because  a  propor- 
tion is  an  equality  between  equal  ratios.     Every  ratio  is 


FUNDAMENTAL  OPERATIONS  53 

a  fraction,  hence  the  equality  symbol  between  the  fractions. 
Formerly  a  proportion  was  written  Ai—bi=A2— 62;  the 
colon  :  and  double  colon  : :  were  used  later,  giving  the  more 
familiar  but  now  obsolete  form  A\  :b\  ::^2:&2-  The  points 
were  later  joined  by  straight  lines  allowing  .A  1/61=^2/62, 
which  in  turn  has  been  replaced  by  the  modern  equation 

form  written  -=—  =  -r— . 
01      62 

A  proportion  contains  four  parts,  viz.,  its  two  numerators 

and  its  two  denominators.     For  convenience  in  referring  to 

these  parts,   the  two  numerators  are   called  antecedents, 

and  the  two  denominators  are  called  consequents,  or  the 

former  are  causes,  and  the  latter  are  effects.     The  first 

numerator  A    and    second    denominator    b,    i.e.,    Ai=z— - 

b2, 

are  called  extremes,  whereas  the  first  denominator  bi  and 

second  numerator  A2,  i.e.,  — =A2,  are  called  the  means. 

bi 

14.  Theorems  on  Proportions.  A  literal  proportion  is 
one  in  which  the  four  parts  or  quantities  are  represented 
by  letters.  When  the  numeric  values  of  these  parts  are 
substituted,  it  becomes  a  numeric  proportion.  When 
letters  and  numerals  are  both  present  it  becomes  a  mixed 
proportion.  The  following  theorems  apply  alike  to  any 
simple  proportion,  i.e.,  a  proportion  of  four  parts.  Con- 
sider the  proportions  representing  the  headings  of  the  first 
five  columns  in  Number  1  of  Table  IV  and  immediately 
under  them  their  altered  form  due  to  the  application  of 
the  theorem  quoted  in  the  column  to  their  right.  If  the 
means  of  a  proportion  are  equal  either  mean  is  called  a 
mean  proportional. 

1st  antecedent _2d  antecedent #     mean  _extreme m 
1st  consequent     2d  consequent'  extreme      mean  ' 

1st  cafuse  _  2d  cause 
1st  effect    2d  effect' 


54 


PRACTICAL  MATHEMATICS 


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FUNDAMENTAL  OPERATIONS 


55 


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56  PRACTICAL  MATHEMATICS 

15.  Projection.  The  projection  of  a  figure  or  object 
is  its  image  or  shadow.  Projection  involves  four  essential 
considerations,  viz.,  an  imaginary  or  real  source  of  illumi- 
nation, such  as  a  lamp  or  the  sun;  a  bundle  or  system  of 
straight  lines,  representing  rays  emanating,  i.e.,  sent  out 
from  the  light  source;  an  object  or  figure  in  the  path 
of  the  rays;  and  lastly  a  screen  or  suitable  surface, 
plane,  or  line  upon  which  to  cast  a  shadow,  image  or 
projection. 

Some  of  the  rays  are  intercepted,  i.e.,  stopped  by  the 
physical  obstruction  of  the  object,  others  pass  around  the 
object,  i.e.,  are  contiguous  to  its  boundary,  and  still  others 
pass  beyond  contact  with  the  object. 

A  shadow  indicates  the  interception  of  the  rays  which 
fall  upon  the  object.  In  obtaining  the  outline  of  the  shadow, 
it  is  sufficient  to  consider  those  rays  which  touch  the  boundary 
of  the  figure  or  object.  Every  point  or  corner  of  the 
object  projects  into  a  point  or  corner  of  the  shadow.     The 

boundary  lines  or  edges  of 
^Njter^-^      A/  the  object  project  into  the 

Xcv^;-c^fY~~~"~--- /'  boundary  lines  or  edges  of 

D  N 0S1  /let  -  C?  ~s£'  f1  the  shadow. 

VvJfcT  -  ^i    /■•'  i  When  the  source  of  illu- 

D  v       v^    i     /  'G  i 

i    X   ^  <,' '">b"         mination  is  a  point  C,  the 

N*V'*S  rays  sPread  into  a  cone  and 

we  have  central  projection 

Fig.  12.— Central  Projection.         as    illustrated    in    Fig.    12. 

The  square  ABDE  has  its 
image  or  projection  at  A'B'D'E'  and  A"B"D"E" '.  The 
points  with  like  accented  letters  are  on  a  single  ray 
showing  clearly  how  a  point  of  the  object  will  project 
along  a  ray  into  a  point  of  the  projection.  In  Fig.  12 
the  object  and  its  images  are  parallel,  and  in  such  cases 
they  will  be  similar  figures.  If  a  screen  is  inserted 
obliquely  to  the  object  the  shadow  or  projection  will  not 
be  a  square. 


FUNDAMENTAL  OPERATIONS  57 

Ex.  86.  From  a  sheet  of  cardboard  cut  a  square  which  will 
be  called  the  die.  The  remaining  part  of  the  cardboard  will 
be  called  a  matrix. 

Place  a  lighted  candle  in  a  darkened  room.  Hold  the  die 
and  then  the  matrix  in  different  positions  before  the  candle  light 
and  measure  the  shadows  on  a  wall.    What  kind  of  figures  result? 

Ex.  87.  Cut  other  dies  in  the  cardboard,  leaving  triangular, 
circular  and  irregular  matrixes  and  repeat  the  detail  of  Ex.  86. 

Ex.  88.  Repeat  Ex.  86,  keeping  the  screen  fixed  but  change 
the  position  of  the  candle. 

Ex.  89.  Place  a  lighted  candle  in  a  cylindrical  glass  vessel 
and  observe  the  projections  of  the  rim  of  the  mouth  on  the  walls 
and  ceiling  as  the  vessel  is  inclined  in  various  positions. 

When  a  matrix  is  placed  before  the  light,  the  rays  diverge, 
i.e.,  the  illumination  spreads  through  the  opening.  The 
area  of  illumination  upon  a  parallel  screen  placed  beyond 
the  matrix  will  vary  as  the  square  of  the  distance  of  the 
former  from  the  light.  Let  A  and  d  represent  the  area 
and  distance  respectively,  of  the  projection,  then  Aozd2. 
In  other  words  if  the  area  is  9  sq.ft.  at  a  distance  of  3  ft., 
then  the  area  will  be  16  sq.ft.  at  a  distance  of  4  ft. 
If  two  areas  A\  and  A2  are  at  the  respective  distances 
di  and  d2,  then 

Ai     di2  A2    d22     !A.      ,  ,.     v 

T2=d?    or    ATd?    (directPr°Portlon)- 

The  intensity  of  illumination,  i.e.,  the  amount  of  light 
falling  upon  each  square  foot  of  screen,  will  decrease  as 
the  areas  increase.  Let  I\  and  I2  be  the  intensity  of 
illumination  on  areas  A\  and  A2,  then 

t:  =  ~t  (inverse  or  indirect  proportion), 
12    Ai 

but  since  -j-  =  -=-«  we  can  substitute  and  obtain 
A!     di2 

h     d22       , . ,.  T     1 

t~  =  -t».  which  means  /  oc  -=, 
I2     di2  d2 


58 


PRACTICAL  MATHEMATICS 


The  intensity  of  illumination  varies  indirectly,  i.e., 
inversely  or  reciprocally  as  the  square  of  the  distance. 

Observation.  In  a  direct  proportion,  the  ratio  of  two 
values  of  one  quantity,  equals  the  direct  or  like  ratio  of  two 
corresponding  values  of  the  other  quantity.  In  an  inverse  or 
indirect  proportion  one  of  these  ratios  is  inverted  or  reciprocated. 
In  central  projection  the  image  or  shadow  is  a  magnifi- 
cation   of    the   object,   die   or  matrix.     The   intensity   of 

any  other  phenomena, 
such  as  sound,  heat, 
electricity,  magnetism, 
which  is  dissipated 
through  space  obeys  the 
same  law  of  the  inverse 
squares. 

When  the  source  of 
illumination  emanates 
parallel  rays  (sunlight 
and  parabolic  reflectors) 
we  have  geometric  pro- 
jection as  illustrated  in 
Fig.  13.  If  the  rays 
meet  the  screen  at  right 
angles  (orthogonal  pro- 
jection) the  boundary  of  the  projection  is  an  exact  duplica- 
tion or  reproduction  in  size  and  shape,  of  the  boundary  of 
the  die.  If  the  rays  are  not  at  right  angles  to  the  screen, 
the  boundary  of  the  projection  is  a  distortion  of  the 
boundary  of  the  die. 

The  shadow  (7'l'l"'3"'3W"7'")  is  obtained  by  parallel 
rays  which  are  oblique  to  the  rear  and  upper  surfaces  of 
the  solid.  The  shadow  (l"2"8"7"6"5"4"3")  is  obtained 
by  the  orthogonal  rays  which  give  a  projection  of  the  upper 
surface. 

Mechanical  drawings  are  geometric  projections  which 
enable  the  workman  to  gain   a  picture  of  a  constructive 


Fig.  13. — Geometric  Projection. 


FUNDAMENTAL  OPERATIONS 


59 


job  as  it  will  appear  from  the  top,  front  and  side  when  com- 
pleted. 

Ex.    90.     Make    mechanical    drawings    after    obtaining   the 
measurements  of  a  rectangular  box;    a  fly-wheel;  and  the  parts 


Fig.  14. — Truncated  Cylinder  and  Spread  Surface. 

of  a  generator.      Show  the  orthogonal  projections  of  the  top, 
front,  and  side  of  the  object. 

Fig.  14  shows  the  mechanical  drawing  of  a  cylinder 
which  has  been  truncated,  i.e.,  cut  obliquely.  Corresponding 
points  and  lines  are  like  numbered  in  each  view.     Viewed 


60  PKACTICAL  MATHEMATICS 

from  above,  the  section,  i.e.,  the  cut  surface  is  a  circle  as 
shown  in  III,  IV,  but  viewed  orthogonally  the  section  I, 
is  shown  in  its  true  shape  as  an  ellipse.  The  dotted  lines 
represent  rays.  The  bottom  or  base  of  the  cylinder  is  a  circle. 
The  lateral  or  vertical  surface  of  an  entire  cylinder  would 
open  into  a  rectangle,  i.e.,  a  rectangular  strip  of  paper 
would  cover  its  lateral  surface.  The  effect  of  truncating 
the  cylinder  is  to  remove  a  portion  of  the  rectangular  sur- 
face as  shown  in  the  development,  i.e.,  spread  surface  V. 
The  development  including  the  base  and  elliptic  section,  VI, 
represents  the  amount  of  material  required  to  be  cut  from 
metal  and  bent  into  shape  in  order  to  cover  the  solid  or 
build  a  like  hollow  vessel. 

The  vertical  lines  in  the  lateral  surface  are  called  elements. 
They  are  parallel  and  equally  spaced,  and  being  twelve 
in  number,  they  intersect  the  base  at  points  which  divide 
the  circumference  into  twelve  equal  arcs.  The  length  of 
the  elements  in  the  development  are  equal  to  the  length 
of  the  corresponding  elements  in  I  and  II. 

Fig.  15  shows  a  cone  with  its  upper  and  lower  half 
tapering  to  the  apex,  0.  The  bases  are  circles.  A  line 
joining  the  apex  with  the  base  is  called  an  element.  The 
development  of  the  lateral  surfaces  will  be  two  sectors 
of  circles  called  nappes,  drawn  with  a  radius  equal  to 
an  element  of  the  cone,  and  a  length  of  arc  equal  to  the 
circumference  of  the  circular  base.  All  sections  parallel 
to  the  base  will  be  circles  as  shown  at  MNPQ. 

A  section  inclined  slightly  non-parallel  to  the  base  is 
an  ellipse,  CDBE. 

A  section  parallel  to  an  element  is  a  parabola,  CFKLG. 

A  section  perpendicular  to  the  base  will  appear  in  both 
halves  of  the  cone.  Every  section  which  cuts  both  halves 
of  the  cone  is  called  an  hyperbola,  KBL  and  HAJ. 

Each  of  the  four  sections  of  the  cone  removes  part  of 
the  conical  surface  which  is  illustrated  in  the  development  V 
and  VI. 


FUNDAMENTAL  OPERATIONS 


61 


Corresponding  points  and  lines  are  like  numbered  and 
lettered  in  each  view. 

Ex.  91.  Make  a  mechanical  drawing  of  a  tin  funnel  and  show 
the  development,  allowing  for  an  edge  for  overlapped  rim  and  an 
extension  for  a  soldered  joint. 

Ex.  92.  Measure  the  shadow  of  a  telegraph  pole  on  a  level 
street.     Let  the  height  of  the  pole  be  represented  by  h,  and  the 


Upper  Nappe 
3__  *  — 5 


K    ■    «        4         3        L 
Lower  Nappe, 


/      i 

J  3""2         1-K12LU  10  7  6      5 

5      6        7         8       9  8      9       10       11  12 


Fig.  15. — Truncated  Cone  and  Lateral  Spread. 

length  of  its  shadow  by  s.  Near  the  pole  insert  a  stick  vertically 
into  the  ground.  Measure  the  height  of  the  stick  and  the  length 
of  its  shadow,  and  designate  these  values  by  fi2  and  S2  respectively. 
If  the  stick  is  driven  deeper  into  the  ground,  the  shadow  shortens 
proportionate!}'.  Therefore  the  shadow  varies  as  the  height  or 
the  height  varies  as  the  length  of  shadow. 


,  ,     h    Si  ,      hz 

hccs     and    t=—     or    hi=—Si 
h2    s2  $2 


or    s  a  h. 


Insert  the  values  obtained  by  measurement  in  the  formula,  and 
express  the  height  of  the  telegraph  pole  at  that  hour  of  that  day. 


62 

s2 


PRACTICAL  MATHEMATICS 


is  a  constant,  and  the  height  of  all  objects  in  the  vicinity  can 


be  quickly  obtained  by  multiplying  the   length  of  an  object's 

shadow  by  — .     Verify  by  experiment  and  make  a  descriptive 

s2 

report. 

Ex.  93.  Repeat  the  experiment  of  Ex.  92  at  a  later  hour  of 
the  day  and  report  your  observations. 

In  mathematics  work  orthogonal  projection  only  is 
used.  Instead  of  projecting  a  line  called  a  projector  upon 
a  surface,  it  is  usually  sufficient  to  project  it  upon  another 
line  called  the  shadow  line,  as  illustrated  in  Fig.  16.     The 


Fig.  I6.7— Orthogonal  Projection. 

rays  are  always  orthogonal  to  the  shadow-line,  and  the 
projection  is  the  segment,  i.e.,  the  portion  of  the  shadow 
line  lying  between  the  extreme  rays. 

Observation.  In  projecting  a  line  it  is  sufficient  to  draw 
the  rays  through  its  extremities.  If  an  extremity  lies  in  the 
shadow-line,  one  ray,  called  a  perpendicular,  drawn  through 
the  other  extremity,  is  sufficient  to  determine  the  projection. 

The  projection  is  equal  in  length  to  the  projector  when 
they  are  parallel,  CD'  and  CD.  If  we  rotate  the  projector 
obliquely  to  the  rays,  the  projection  is  shortened,  EF  and  PR. 
In  the  extreme  position,  BB'  with  the  projector  parallel  to 


FUNDAMENTAL  OPERATIONS 


63 


the   rays,  the  projection  is  a  point.     When  the  projector 
rotates  it  is  called  a  radius  vector  (R.V.),  or  moving  arm. 

Ex.  94.  Make  a  list  of  projectors  and  their  corresponding 
projections  in  Fig.  16,  giving  their  figure  names. 

Ex.  95.  In  Fig.  16  state  the  vertical  and  horizontal  projec- 
tions of  the  lines  passing  through  the  point  1,0;  7,  4;  12,  14. 


-- gijiv ■* 

Fig.  17. — Angles  of  Four  Quadrants. 

16.  Trigonometric  Functions.  Fig.  17  represents  two 
perpendicular  diameters  which  divide  the  circle  into 
four  equal  quarters  called  quadrants,  I,  II,  III,  IV.     The 


64  PRACTICAL  MATHEMATICS 

numbers  of  the  quadrants  begin  at  the  upper  right-hand 
quarter  or  corner  and  are  ordered  in  a  counter-clockwise 
direction,  i.e.,  opposite  to  the  motion  of  clock  hands.  The 
angles  1,  2,  3,  and  4  are  called  respectively  angles  of 
the  first,  second,  third  and  fourth  quadrants,  because 
their  terminal  sides,  i.e.,  radii  vectors,  lie  in  these  respective 
quadrants.  Their  initial  sides  are  the  same  horizontal 
line  and  their  common  vertex  is  the  center  of  the  circle. 
Angles  1,  2,  3,  4  are  represented  above  the  circle 
separately,  with  their  sides  prolonged  to  facilitate  measure- 
ment. A  number  of  lines  or  rays  are  drawn  from  the 
terminal  sides  perpendicular  to  the  initial  side  or  the  initial 
side  produced.  A  radius  vector  may  be  considered  as  any 
portion  of  the  terminal  side  measured  from  the  vertex 
to  the  beginning  of  a  ray  or  perpendicular.  The  inter- 
section of  the  perpendicular  with  the  initial  side  is  called 
its  foot.  The  distance  between  the  foot  of  the  perpen- 
dicular and  the  vertex  is  the  projection  of  that  particular 
radius  vector.  By  these  rays  we  have  formed  four  groups 
of  similar  figures. 

Measure  and  record  the  length  of  a  radius  vector 
OC,  its  corresponding  perpendicular  EC,  and  projection 
OE  for  angle  1. 

From  the  three  measured  lines  it  will  be  possible  to 
write  three  distinct  ratios  and  their  three  reciprocals.  In 
terms  of  radius  vector,  perpendicular  and  projection  they 
are  named  as  follows: 

The  sine  of  Z  1  (abbreviated  sin  1) 

_  perpendicular  _    J_    _  EC 
radius  vector     R.V.     OC ' 

The  cosine  of  Z 1  (abbreviated  cos  1) 

proj  ection     _  Pro  j .  _  OE 
radius  vector    R.V      OC 

The  tangent  of  Z  1  (abbreviated  tan  I) 

_  perpendicular  _    J_    _  EC^ 
projection        Proj .     OE' 


FUNDAMENTAL  OPERATIONS  65 

The  reciprocal  of  a  sine  is  called  a  cosecant  (esc). 
The  reciprocal  of  a  cosine  is  called  a  secant  (sec). 
The  reciprocal  of  a  tangent  is  called  a  cotangent  (cot.). 

- — 7— =  cscZl:        — 7— =  secZl: —  =  cotZl. 

sin/1  cos  Z  1  tan  Z  1 

- — — -  =  sinZl:     — -  =  cosZl:     — .    .  .,  =tanZl. 

esc  Z  1  sec  Z  1  cot  Z  1 

These  six  ratios  are  called  trigonometric  functions, 
because  their  numeric  values  depend  upon  the  magnitude 
of  the  angle  under  observation.  ,Any  distance,  such  as 
OA  or  OD,  may  be  used  as  the  radius  vector  provided 
it  is  measured  outward  from  the  vertex.  For  every  such 
radius  vector,  there  will  be  a  corresponding  perpendicular 
such  as  AB  or  DF,  and  in  like  manner  a  corresponding 
projection  OB  or  OF. 

Ex.  96.  Determine  the  sine,  cosine,  and  tangent  of  angle  1, 
using  two  new  sets  of  radii  vectors  with  their  corresponding  per- 
pendiculars and  projections.  The  ratios  obtained  for  each  function 
should  be  numerically  the  same,  as  the  three  measured  parts 
form  similar  triangles  in  each  instance.  Why  does  this  make  the 
respective  ratios  constant  for  angle  1? 

Repeating  the  determination  of  the  values  of  functions 
for  different  angles,  leads  us  to  the  observation,  that  each 
function  of  a  given  angle  is  a  numeric  constant,  i.e.,  a 
definite  fixed  number. 

17.  The  Use  of  Trigonometric  Tables.  The  numeric 
values  of  the  trigonometric  functions  of  acute  angles  are 
given  in  Table  VIII.  Reading  from  the  top  down,  the  first 
column  gives  the  number  of  degrees  of  angles  from  0° 
to  45°.  The  fourth,  fifth,  sixth,  and  seventh  columns 
give  respectively  the  sine,  tangent,  cotangent,  and  cosine 
of  the  angles,  and  are  to  be  read  on  the  same  horizontal 
line  with  the  corresponding  angle. 


66  PRACTICAL  MATHEMATICS 

Ex.  97.  Obtain  the  sine  of  20°  from  the  table.  Locate  20 
in  the  degree  column,  carry  the  finger  horizontally  across  the 
page  and  in  the  column  headed  sin  read  the  decimal  .3420.  If 
a  card  is  placed  horizontally  under  20  it  will  assist  the  eye  in  keeping 
to  the  horizontal  line. 

Ex.  98.  Obtain  the  tangent  of  38°.  In  the  degree  column 
locate  38  and  horizontally  across  and  in  the  tangent  column 
read  .7813. 

Ex.  99.  Obtain  the  cotangent  of  9°.  Look  up  9  in  the  degree 
column,  and  horizontally  across  and  in  the  cotangent  column 
read  6.3138.* 

Ex.  100.     Obtain  the  cosine  of  41°,  41.5°,  and  42°. 
cos  41°  =  .7547 
cos  42°  =.7431 


.0116  =  decrease  for  1°  between  41°  and  42°. 
.0116 X. 5  =.0058=  decrease  for  .5°  between  41°  and  42°. 
/.    cos  41 .5°  =  .7431  +.0058  =  .7489  =  .7547  -  .0058. 

Observation.  To  interpolate,  i.e.,  make  the  correction  for 
a  decimal  angle,  multiply  the  decimal  excess  by  the  difference 
in  the  numeric  values  of  the  same  function  of  the  next  lower 
and  next  higher  angle.  Add  the  correction  for  sines  and 
tangents,  because  sines  and  tangents  increase  with  increasing 
angles.  Subtract  the  correction  for  cosines  and  cotangents, 
because  cosines  and  cotangents  decrease  with  increasing  angles. 

Ex.  101.  Obtain  the  sines  of  30°,  27°,  32.3°,  44°,  8.8°;  the 
cosines  of  36.2°,  30°,  18°,  23.7°,  1.5°;  the  tangents  of  30°,  45°, 
15.8°,  20°,  2°;  the  cotangents  30°,  16°,  7°,  42.5°,  3°,  15°. 

A  further  examination  of  Table  VIII  shows  that  functions 
of  acute  angles  from  45°  to  90°  are  read  from  the  bottom 
upward.  Instead  of  reading  the  headings  of  the  columns, 
we  read  the  footings,  i.e.,  the  designation  at  the  bottom 
of  the  columns.  This  arises  from  the  fact  that  the  numeric 
value  of  the  function  of  an  angle  is  also  the  numeric 
value  of  the  cof unction  of  the  complementary  angle.  Sines 
and  cosines  are  cof  unctions,  and  so  are  tangents  and  cotan- 
gents,  also   secants   and   cosecants.    Thus  the   sine   60= 


FUNDAMENTAL  OPERATIONS  67 

cosine  30  =  .8660;  tan  28°  =  cot  62°  =.5317.  To  read  func- 
tions of  an  acute  angle  greater  than  45°,  look  up  the  angle 
in  the  degree  column,  carry  the  finger  horizontally  to  the 
left  until  within  the  column  with  the  function  name  at  the 
bottom.  Thus  the  sin  77°  =  .9744;  tan  65°  =  2.1445;  cos 
53.5°  =.5948. 


Ex.  102.  Obtain  the  sine,  cosine,  tangent  and  cotangent  of 
the  following  angles:  57.3°,  87.6°,  47.6°,  72.9°. 

Ex.  103.  What  is  the  angle  whose  sin  is  (.8660)?  In  order 
to  obtain  the  angle  the  table  is  used  inversely,  i.e.,  in  a 
reverse  manner,  and  therefore  the  angle  is  called  the  inverse  or 
antifunction.  It  is  customary  to  abbreviate  the  angle  by  some 
Greek  letter  such  as  6  (theta)  and  the  words  "  whose  sin  is  " 
by  writing  (  —  1)  above  the  function  name.  Using  symbols  the 
example  may  be  stated  ^=sin_1  (.8660).  In  the  seventh  column 
we  find  .8660,  with  the  word  sine  at  the  foot  of  the  column.  This 
means  we  read  the  angle  from  the  bottom,  and  therefore  in  the 
last  column  we  find  60°  on  the  same  horizontal  line  with  .8660. 
Therefore  0=60°. 

Ex.  104.  <£=tan-1  (2.1445),  which  means  <£  is  the  angle 
whose  tangent  is  (2.1445).    What  is  the  degree  measure  of  <£? 

Ex.  105.  a  =  cos-1  (.5446),  which  means  a  is  the  angle  whose 
cosine  is  (.5446).  What  is  the  radian  measure  of  a?  Instead 
of  reading  the  angle  in  degrees  in  the  first  column,  read  its  radian 
measure  on  the  same  horizontal  line  of  the  second  column.  One 
radian  of  angle  equals  57.3°  approximately.  Since  there  are  2?r 
radians  to  every  360°  of  angle,  then,  one  degree  equals  .0175 
radian  approximately. 

18.  Functions  of  Angles  of  Any  Quadrant.  To  obtain 
the  numeric  value  of  the  function  of  an  angle  greater  than 
90°,  we  use  Table  VIII. 

The  functions  of  angles  of  any  quadrant  are  ratios, 
and  are  defined  in  exactly  the  same  way  as  functions  of 
acute  angles.  They  may  be  obtained  by  drawing  and 
measuring  radii  vectors,  and  their  corresponding  perpen- 
diculars and  projections  as  shown  in  Fig.  17,  and  expressing 
these  as  ratios.     These  ratios  vary  through  the  same  range 


68  PRACTICAL  MATHEMATICS 

of  numeric  values  given  in  Table  VIII.  Therefore,  in  order 
to  use  Table  VIII  to  determine  the  functions  of  an  obtuse 
angle,  look  up,  i.e.,  obtain  the  functions  of  its  supplement. 
In  order  to  obtain  the  functions  of  a  III  quadrant  angle 
subtract  180°  from  the  angle  and  look  up  the  functions  of 
the  remainder.  In  order  to  obtain  the  functions  of  a  IV 
quadrant  angle  subtract  the  angle  from  360°  and  look  up 
the  functions  of  the  remainder. 

Thus  the  sine  of  137°  equals  the  sine  of  its  supplement. 

sin  137°  m sin(180° - 137°)  =  sin  43°  =  .6820. 

The  tan  212°=tan  (212°  - 180 °)  =  tan  32°  =  .6249. 

The  cos  314°  =  cos  (360° -314°)=  cos  46°  =  .6947. 

In  any  case  the  angle  is  combined  by  subtraction  with 
180°  or  a  multiple  of  180°. 

There  is  this  slight  difference,  however,  that  we  must 
observe  a  convention  of  signs  in  measuring  perpendiculars 
and  projections,  i.e.,  vertical  and  horizontal  lines  as  shoAvn 
in  Fig.  17.     Accordingly: 

A  perpendicular  is  positive  if  it  extends  above  the 
initial  side  or  the  initial  side  produced  to  the  left  of  the 
vertex,  i.e.,  it  must  be  above  the  projection,  EC  and  HL. 
It  is  negative  if  it  extends  below  the  projection,  PN  and  SA. 

A  projection  is  positive  if  it  extends  to  the  right  of 
the  vertex,  OF  and  OV.  It  is  negative  if  it  extends  to 
the  left  of  the  vertex,  0 J  and  OR . 

Since  the  radius  vector  is  a  moving  arm,  it  is  a  line 
without  definite  direction.  It  is  regarded  without  sign, 
which  is  equivalent  to  saying  it  is  positive  in  sign. 

In  consequence  of  the  convention  of  signs,  all  functions 
of  acute  angles  are  positive.  In  the  other  three  quadrants, 
some  of  these  ratios  will  be  negative,  i.e.,  when  either 
their  numerators  or  their  denominators  are  negative.  In 
any  quadrant  these  ratios  will  be  positive  if  numerator  and 
denominator  have  like  signs,  i.e.,  both  positive  or  both 
negative.  There  will  be  cycles  of  change  in  the  signs  of 
the  functions  as  shown  in  Table  V. 


FUNDAMENTAL  OPERATIONS 


69 


Ex.  106.  Verify  the  following  Table  V  for  the  signs  of 
functions  of  angles  of  quadrants  I,  II,  III,  and  IV,  by  observing 
the  convention  of  signs. 


TABLE  V.     SIGNS  OF  QUADRANTS 


Functions. 

Quadrants. 

I 

II                       III 

IV 

sin,  esc  

+                        + 

- 

tan,  cot 

+ 

+ 

- 

cos,  sec 

+ 

- 

+ 

Ex.  107.     The  sin  212°  =  -sin  32°  =  -.5299. 

The  cos  137°  =  -cos  43°  -  - .7314.     Why? 

The  tan  314°  =  -tan  46°  =  - 1.0355.     Why? 
Write  the  values  of  the  following:   tan  263°;   sin  281 
cos  300°:  sin  300°;  tan  300°. 


Why? 


cos  157c 


In  Table  VIII  the  numeric  value  of  cot  0°  and  tan 
90°  is  not  written,  but  is  expressed  by  the  symbol  °° , 
called  infinity.  This  means  that  the  numeric  value  is 
so  excessively  large  as  to  be  inexpressible  with  figures. 
Secants  and  cosecants  are  not  given,  as  they  are  more  easily 
replaced  by  their  respective  reciprocals  cosines  and  sines. 

Ex.  108.  Look  up  the  values  of  the  sine  and  the  cosine  of  30°. 
Divide  the  former  by  the  latter,  and  observe  their  quotient  in  the 
tangent  column  on  the  same  line.  Perform  this  division  for 
angles  of  15°,  20°,  25°,  35°,  40°,  45°.  The  last  or  fourth  figure 
of  Table  V  is  not  dependable,  so  there  may  be  a  slight  variance 
in  the  fourth  figure  of  the  quotient. 

Observation.  The  tangent  of  an  angle  is  the  ratio  of  its 
sine  to  its  cosine. 

Ex.  109.  Construct  a  right  triangle.  Label  the  right  angle 
C  (a  conventional  method)  and  the  acute  angles  A  and  B  respect- 
ively. At  the  middle  of  each  side  place  a  lower  case  letter  corre- 
sponding to  the  capital  letter  at  the  opposite  vertex. 


70  PRACTICAL  MATHEMATICS 

Formulate,  i.e.,  write,  the  names  of  the  functions  of  angles 

A   and  B,  connecting  them  with  an  equal  sign  to  the  following 

a    a     b     b 
ratios:  — ;  — ;  — :  — . 
coca 

19.  The  Right  Triangle.  Each  of  these  four  statements 
contains  three  quantities,  viz.,  two  sides  and  a  function  of 
an  angle.  If  any  two  of  these  three  quantities  are  known, 
the  third  may  be  obtained  by  substituting  the  numeric 
values  of  the  former  in  the  proper  equation.  The  simplified 
result  will  be  the  value  of  the  latter.  If  each  of  the  remain- 
ing parts  of  the  triangle  has  its  numeric  value  deter- 
mined in  this  way,  the  triangle  is  said  to  be  solved  by 
trigonometry. 

Ex.  110.    A  right  triangle  has  a  side  a  =4.33",  and  its  hypote- 

a 


nuse  c  =5."    Determine  angles  A  and  B,  and  the  side  b.    Sin  A  = 
.8660.    Referring  to  Table  VIII  under  the  sin  column  for 


c 
4.33 
5 

.8660  we  locate  it  horizontally  across  from  angle  60°.  Therefore 
A  =60°,  but  C  =90°,  and  since  the  sum  of  the  angles  of  a  triangle 
equals  180°,  then  A  and  B  are  complementary.  Therefore 
B  =90°  -A  =90°  -60°  =30°.    b  may  be  obtained  by  substituting 

in  the  tangent    formula    tan  A  =-,   or  in  the  cosine  formula 

cos  A  =— .    Using  the  latter  we    have  c  cos  A  =  6  (Mul.  ax.). 
c 

Therefore  substituting  for  c  and  COB  A,  we  have  b  =30  X  cos  60° 
=30X.5  =  15". 

Ex.  111.  A  right  triangle  has  angle  A  =30°  and  hypotenuse 
c=20". 

Solution: 

sin  A  =-,  therefore  a  =  c  sin  A  =20  X.5  =  10";  A  =30° 

c  c  =  20" 


b  a  =  1°" 

cos  A  =-,  therefore  b  =c  cos  A  =20  X. 866  =  17.32";    b  =  17.32' 
c  B  =60 


£=90°-A=90°-30°=60c 


FUNDAMENTAL  OPERATIONS  71 

Ex.  112.    A  right  triangle  has  an  angle  A  =29°  and  side  a  =380. 
a 

b 


tan  A  =—.     .'.  b  tan  A  =a  (Mul.  ax.).    The  multiplier  of  a  func 


tion  precedes  the  function.    By  applying  division  axiom,  6  =  - : 

tan  A 

380 
=  — —  =  685.    The  value  of  c  may  be  obtained  by  using  the  formula 

.o04 

sin  A  =— ,  from  which  we  obtain  c  sin  A  =a,  and  therefore  c  =- — - 
c  sin  A 

=  -j~=785;  B  =90° -29°  =61°. 

Observation.  The  number  of  dependable  figures  in  a 
quotient  is  the  same  as  the  least  number  of  dependable  figures 
in  either  the  numerator  or  denominator  of  the  corresponding 
fraction. 

Ex.  113.  Construct  the  following  right  triangles  and  solve 
them  graphically,  i.e.,  by  measuring  the  sides  and  angles.  In 
order  to  obtain  graphic  results  which  are  comparable  with  cal- 
culated ones,  it  is  necessary  that  the  figure  should  be  drawn 
with  hair  lines,  by  using  a  hard  pencil  and  then  making  all 
measurements  with  accuracy  before  inking  the  figure.  Determine 
the  solution  by  calculation,  using  trigonometric  formulas: 

(1)  a  =4        6=3  (4)  c=26    6=24 

(2)  a  =  15       c=25  (5)  c=39    6  =  15 

(3)  c  =  12.5    6  =  10  (6)  a  =  12     6=5 

If  the  work  is  performed  carefully,  it  will  be  observed 
that  triangles  (1),  (2),  (3)  are  a  group  of  similar  triangles, 
in  which  the  sides  have  the  ratio  5:4:3.  Triangles  (4),  (5), 
(6)  form  another  group  of  similar  triangles  in  which  the 
sides  have  the  ratio  13:12:5.  There  are  very,  few  right 
triangles  of  the  type  where  the  sides  and  hypotenuse  are 
integral  and  commensurable,  i.e.,  have  a  common  unit  of 
measure.  The  3-4-5  triangle  is  well  known  by  builders, 
who  use  it  in  constructing  right-angled  framework.     It  is 


72 


PRACTICAL  MATHEMATICS 


used  also  for  staking  out  rough  surveys  in  which  the  courses, 
i.e.,  the  boundaries,  are  at  right  angles. 

In  each  one  of  the  above  examples,  square  the  three 
sides  of  each  triangle,  and  show  that  in  a  right  triangle 
the  square  of  the  hypotenuse  equals  the  sum  of  the 
squares  of  the  two  legs. 

This  theorem  may  be  recognized  from  the  construction 
shown  in  Fig.  18.     It  is  due  to  its  constructive  proof  that 


J  H 

Fig.  18. — Squares  on  a  Right  Triangle. 

it  is  known  after  a  Greek  philosopher  as  the  Pythagorean 
Theorem.  Construct  a  right  triangle  on  squared  paper. 
Using  the  hypotenuse  and  each  leg  as  respective  sides, 
carefully  construct  squares  on  these  sides  as  shown  in 
Fig.  18.  Count  the  small  squares  to  verify  the  theorem. 
Note  the  division  of  the  largest  square  into  trapezoids 
and  an  included  square.  Identify  the  trapezoids  with  the 
divisions  of  the  square  constructed  on  the  longer  of  the 
two  legs.    The  trapezoids  include  a  square  equal  to  the 


FUNDAMENTAL  OPERATIONS  73 

square  constructed  on  the  shorter  leg,  The  formulation 
of  this  theorem  in  terms  of  hypotenuse  c  and  legs  a  and  b 
is  expressed  as: 


c2  =  a2+b2    or    c  =  Va2+62     (Root  Ax.); 
letc  =  5,  6  =  4,  a  =  3,  then; 


25  =  9  +  16     or     5  =  V9+16  =  ^25. 


(1)  a=6 

c  =  10 

(2)  6=7 

c  =  12 

(3)  a  =5 

.    6=6 

a2  =  c2_52.  a  =  Vc2-62;  62  =  c2-a2;  b  =  V(?-a2. 

Ex.  114.  Determine  the  third  side,  given  the  following  data 
for  right  triangles: 

(4)  a=3.2    c=5.4 

(5)  6=3.5    c=7 

(6)  o  =  150    6  =  175 

20.  Square    Root.      To  obtain  the  square   root  of    a 

number,  such  as  600.25,  proceed  as  follows: 

1 .  To  the  left  and  right  of,  and  beginning  at  the  decimal 
point,  place  the  figures  in  groups  of  two.     6  W  .25' . 

2.  Under  the  extreme  left-hand  group  (6),  write  and 
subtract  the  nearest  squared  number  (4),  and  place  (2) 
the  root  of  (4),  as  the  first  figure  of  the  quotient. 


6W.25' 
4 


2  quotient 


40  trial  divisor. 
2  00  4  second  term  root 

44  complete  divisor. 

3.  Bring  down  the  next  group  (00)  alongside  the  re- 
mainder (2)  to  form  the  new  dividend  (200). 

4.  The  trial  divisor  (40)  is  twenty  times  the  root  obtained 
so  far. 

5.  Divide  the  trial  divisor  (40)  into  the  new  dividend 
(200),  and  obtain  the  next,  i.e.,  second  figure  (4)  of  the 
quotient,  and  also  add  the  same  (4)  to  complete  the  divisor. 


74  PKACTICAL  MATHEMATICS 

6.  From  the  new  dividend,  subtract  (176)  the  product 
of  the  second  figure  (4)  of  the  quotient  times  the  complete 
divisor  (44) 

7.  Bring  down  the  next  group  alongside  the  last  remainder. 
Continue  the  process  by  repeating  in  order  4,  5,  6,  7, 

until  the  groups  are  exhausted.  The  root  has  as  many 
integral  and  decimal  figures,  respectively,  as  the  given 
number  has  integral  and  decimal  groups.  A  zero  may  be 
written  after  the  last  decimal  figure  to  complete  a  decimal 
group,  if  an  odd  number  of  decimal  figures  are  given. 

2    4.5    Root 

2X20  =  40,  twenty  times  root  obtained  so  far 
40  trial  divisor 
+  4  second  term  root 
44  complete  divisor  (44X4  =  176) 
24  X  20  =  480,  twenty  times  root  obtained  so  far 
480  =  trial  divisor 
+  5  =  third  term  root 
485  =  complete  divisor  (485X5  =  2425). 

/.     V60O25  =  24.5. 

21.  Operations  upon  Fractions. 

Ex.  115.  An  electric  circuit  contains  a  battery  which  has 
an  electromotive  force  of  10.36  volts.  The  resistance  of  the  entire 
circuit  is  93.2  ohms.  What  current  in  amperes  is  flowing  through 
the  circuit? 

By  Ohm's  Law,  the  intensity  (I)  of  current  in  amperes 
which  flows  through  a  circuit,  equals  the  ratio  of  the 
voltage  (E)  applied,  i.e.,  put  in  the  circuit  to  the  resistance 
(R)  of  the  circuit. 

This  law  is  formulated,  i.e.,  expressed  by  a  formula  as 

follows : 

/i  \  rpi  +  /  x     electromotive  force  (volts) 

(1)  The  current  (amperes)  = v— ,—y — \ -. 

resistance  (ohms) 


6W.'25 

4 

2  00 

176 
24    25 
24    25 


FUNDAMENTAL  OPERATIONS  75 

Using  the  notation,  i.e.,  the  symbols  which  abbreviate 
current,  electromotive  force,  and  resistance,  to  replace  these 
words  in  equation  (1)  we  have  (2) : 

(2)  I  =  | 

This  is  a  working  formula,  and  therefore  is  an  equation, 
i.e.,  an  expression  of  equality  between  quantities. 

The  data,  i.e.,  the  numeric  values  of  quantities  given 
in  the  example,  are  substitued  in  (2),  and  since  J57=  10.36, 
and  #  =  93.2,  we  obtain  (3) : 

'-» 

Performing   the   indicated    division  -^— =  0.11-f,    the 

ao.2 

value  of  the  current  is  expressed  in  (4). 
(4)  7  =  0.11  amperes. 

The  answer  is  indicated  not  by  writing  Ans.  or  Res., 
but  by  underscoring  and  writing  the  meaning  of  the  answer, 
and  the  unit  of  measure,  after  the  numeric  value  of  the 
solution. 

The  equality  sign  separates  the  equation  into  two  parts, 
called  respectively  the  left-hand,  or  first  member,  and  the 
right-hand  or  second  member. 

Ex.  116.  A  dynamo  supplies  a  current  at  115  volts,  to  a  circuit 
containing  a  resistance  of  220  ohms.  What  is  the  value  of  the 
current  in  amperes  flowing  through  the  circuit?  Use  equation 
(2). 

Ex.  117.  A  dynamo  supplies  a  current  of  115  volts,  to  a  circuit 
containing  a  resistance  of  110  ohms.  What  is  the  value  of  the 
current  in  amperes  flowing  through  the  circuit? 

Observation.  How  does  the  value  of  a  fraction  change, 
when  the  value  in  its  numerator  is  kept  constant,  but  the  value 
in  the  denominator  is  decreased? 


76  PRACTICAL  MATHEMATICS 

How  does  the  value  of  a  fraction  change,  when  the  value  in 
the  numerator  is  kept  constant,  but  the  value  in  the  denom- 
inator is  increased? 

Ex.  118.  A  battery  of  storage  cells  supplies  a  current  of  50 
volts  E.M.F.  to  a  circuit  having  a  resistance  of  220  volts.  What 
is  the  value  of  the  current  flowing  through  the  circuit?  Compare 
with  the  result  of  Ex.  116. 

Observation.  How  does  the  value  of  a  fraction  change, 
when  the  numeric  value  of  its  denominator  is  kept  constant, 
but  the  value  of  the  numerator  is  decreased? 

Observation.  How  does  the  value  of  a  fraction  change, 
when  the  numeric  value  of  its  denominator  is  kept  constant, 
but  the  value  of  the  numerator  is  increased? 

Ex.  119.  A  resistance  of  2420  ohms  is  connected  to  the  over- 
head wire  and  also  the  track  of  a  trolley  line.  Between  the  trolley 
wire  and  track  is  a  voltage  difference  (P.D.)  of  550.  What 
current  will  flow  through  the  resistance?  Compare  with  the  result 
of  Ex.  118. 

Ex.  120.  A  battery  of  10  volts,  E.M.  F.  is  attached  to  complete 
a  circuit  of  44  ohms  resistance.  What  current  passes,  i.e.,  flows 
through  the  circuit?     Compare  with  the  result  of  Exs.  118  and  119. 

Observation.  What  is  the  effect  upon  the  numeric  value 
of  a  fraction  when  its  numerator  and  denominator  are  either 
both  multiplied  or  both  divided  by  the  same  factor?  This  state- 
ment is  known  as  the  axiom  of  fractions,  and  is  abbreviated  by 
ax.  frac. 

It  is  the  axiom  of  fractions  which  is  applied  in  simpli- 
fying, reducing,  and  uniting  fractions,  in  the  operations  of 
addition,  subtraction,  multiplication,  and  division. 

Ex.  121.  Draw  a  unit  square  I  (one  inch  on  a  side).  Join 
the  middle  of  the  upper  side  to  the  middle  of  the  lower  side.  The 
unit  square  is  divided  into  two  equal  parts,  i.e.,  rectangles.  One 
of  these  parts  equals  a  unit  divided  by  2,  i.e.,  \.  The  fraction  ^ 
is  made  up  of  the  numerator  1,  which  stands  for  the  whole  unit, 
i.e.,  the  magnitude  of  the  unit  square.  Under  it  is  a  dividing 
line  indicating  a  division  of  the  unit,  and  beneath  the  latter  is  the 


FUNDAMENTAL  OPERATIONS  77 

2  which  indicates  the  number  of  parts.  If  the  unit  square  were 
divided  into  three,  four,  five  or  six  equal  rectangles,  each  of  the 
parts  of  the  unit  would  be  represented  by  the  respective  fractions 
h  I,  i,  or  |.  If  the  unit  square  is  cut  apart  along  the  lines  of 
division,  and  the  parts  reassembled  to  produce  any  other  figure, 
the  area  of  the  latter  would  be  one  unit.  Why?  Therefore  a 
unit  of  area  of  any  shape,  such  as  a  unit  circle,  i.e.,  a  circle  of 
unit  radius  when  divided  into  two,  three,  four,  five,  six  parts  will 
have  these  parts  represented  by  the  fractional  units,  i.e.,  the 
fractions  },  |,  \,  |,  \. 

Ex.  122.  Construct  a  square  I  of  1  unit  area,  another  square 
II  of  2  units  area,  another  square  III  of  3  units  area.  Divide 
each  square  into  6  parts.  How  many  parts  of  I  equal  one  part 
of  II? 

1      1      ,  «    2 
•'•     2timeS6"=6    °f  2=6- 


To  get  equal  divisions  the  two-unit  area  II  would  be  divided 
into  6  parts  and  the  one-unit  area  I  would  be  divided  into  3  parts. 
In  other  words  f  =  \. 

How  many  parts  of  I  equal  one  part  of  III? 


11  3      1 

3  times -=-  of  3=-=- 

6     6  6     2 


Ex.  123.  Construct  a  square.  Call  its  area  a  units.  Divide 
the  square  into  two,  three,  four,  five  six,  equal  rectangles.     These 

parts  are  designated  by  — ,  — ,  — ,  — ,  — ,  respectively. 
z     6     4     5     o 

Interpret    each    of    the    following    fractions    in    two    ways: 

L  £  £  L  1L  M 

2'  4'  8'  1.6'  32'  64' 

Write  the  fractions  which  correspond  to  the  divisions  of  a  unit 
area  into  a,  b,  c,  d,  e  parts  respectively. 

Ex.  124.  Construct  a  square  of  K  units,  and  divide  it  into  m 
equal  rectangles.  What  fraction  represents  the  area  of  one  of 
its  rectangles? 

Ex.  125.    What  is  the  interpretation  of  the  following  fractions: 
a    c     d    5     3a    2c    d     5     3a 
b'  3 '  V'  /'  ~b'  3~'  2e'  3f  26' 


78  PRACTICAL  MATHEMATICS 

Ex.  126.    Arrange  the  fractions  -,  -,  -,  — ,  — ,  —  according 

to  the  order  of  their  magnitudes?  If  we  interpret  these  fractions 
as  multiple  values  of  different  subdivisions  of  a  unit  area,  then 
the  smallest  division  has  been  obtained  by  dividing  the  unit  area 
into  64  rectangles.  Two  of  these  64  rectangles  would  unite  into 
a  larger  rectangle,  which  could  be  obtained  by  dividing  the  unit 

2      1 

area  into  32  rectangles.     In  other  words,  —  =— .    In  like  manner 

o4     62, 

2       1  4       1 

our  reasoning  would  show  us  that  —  =—  and  — ■  =— ;  further, 

62     lo  o4     lb 

vA=k=k>   further   !44=I=i   «^d  further  |-| 

4     ]T  16    32 
"8  "16  "32  "64" 


All  these  facts  verify  the  axiom  of  fractions,  viz.: 
If  the  numerator  and  denominator  of  a  fraction  are  either 
both  multiplied  or  both  divided  by  a  like  factor,  the  fraction 
remains  unaltered  in  value,  although  it  is  changed  in  form. 
By  applying  the  axiom  of  fractions,  we  have : 


1 

1X32 

32 

2 

2X32 

64 ' 

3 

3X16 

48 

4 

4X16 

64' 

5 

5X8 

40 

8 

■  8X8 

64' 

7 

7X4 

28 

16 

16X4 

64' 

9 

9X2 

18 

32 

32X2 

64' 

11 

11X1 

11 

64 

64X1 

"64' 

FUNDAMENTAL  OPERATIONS  79 

In  the  order  of  ascending,  i.e.,  increasing  magnitudes, 
these  fractions  should  be  written  as  follows: 

II  JL  L  I  *L  §_ 

64'   32'    16'    2'    8'    4* 


Literal  and  mixed  (literal  and  numeric  combined) 
fractions  are  treated  alike  according  to  the  laws  for  adding, 
subtracting,  multiplying,  and  dividing  numeric  fractions. 

Ex.  127.    Unite  -  +-  +j-  +—  --.    The  first  step  is  to  apply 

the  axiom  of  fractions,   in  order  to  have  a  like  denominator  for 
each  fraction.    Therefore  the  example  becomes 

10    28     o^    26_12 
16+16+16+16    16' 

In  other  words  we  have  (10  +28  +a  +26  — 12)  of  the  same  parts, 

a          •     *            10 +28 +a +26 -12   . 
i.e.,  common  denominator,  or  — ,  by  combining 

numerators   and  writing   their    sum   over  the    common    denom- 
inator.    This   simplifies   and   reduces   to    — .    Since    the 

lb 


literal  and  numeric  terms  cannot  be  united,  their  sum  is  indi- 
cated by  the  plus  sign  connecting  them. 

Ex.  128.    Unite    f +^+^+t|;-A+2-      In    order     to 
8a    46     16a     166    4a6 

obtain  a  common  denominator  for  the  given  fractions,  we  observe 
the  different  factors  in  each  denominator.  These  factors  must 
be  contained  in  the  least  common  denominator  (L.C.D.).  Factor- 
ing the  denominators,  we  have: 

8a=2x2x2xa, 

46=2X2X6, 
16a=2x2x2x2Xa, 
166=2X2X2X2X6, 
4a6=2x2XaX6, 
16ab  =2  X2  X2  X2  Xa  Xb  =L.C.D. 


80  PRACTICAL  MATHEMATICS 

2 

The  last  term  in  the  given  sum  may  be  written  —  instead  of  2. 

Each  fraction  of  the  example  is  multiplied  by  the  deficient,  i.e., 
lacking  factors  in  its  deniminator,  hence : 

5x26      7  x4a        1  X6        2Xa  _  3x4      2Xl6a6 
8oX26+46x4a+16aX6    166  Xa    4a6x4     Ixl6a6 

_  }°]L      28a     _6_     J2a_ 12^     32a6 

"i6a6+i6a6+16a6+16a6~  16a6+16a6" 

Since  each  fraction  now  appears  with  a  like  denominator,  we  may 

collect  the  numerators,  and  write  their  sum  with  the  same  divisor 

,.  ,     .       116+30a+32a6-12 
or  common  denominator,  which  gives rr-r . 


When  fractions  appear  in    an    equation,  they  may  be 
removed  by  multiplying  the  equation  by  the  L.C.D. 

Ex.  129. 

(1)  f+J-*Tp the  L.C.D.  =12. 

(2)  52p  +— p  =  12X0;  -^- ,-by  multiplying  (1)  by  L.C.D. 

simplifying  (2). 
subtracting  12z  from  (3). 
subtracting  20  from  (4). 

dividing  (5)  by  6. 


The  operation  in  (4)  is  often  called  transposition,  because 
a  term  appears  to  be  carried  across  the  equality  sign  with 
a  change  in  its  sign. 

In  multiplying  fractions,  write  the  answer  as  a  fraction, 
with  a  numerator  equal  to  the  product  of  the  individual 
numerators,  and  with  a  denominator  equal  to  the  product 
of  the  individual  denominators. 


(3)  20+18x  =  12x-l 

4, 

(4)  20+6x=-14,  - 

(5)          6x=-34,  - 

m      —?= 

17 
3' 

FUNDAMENTAL  OPERATIONS  81 

t.     < on    5     3     6     14    5X3X6X14    ,    .    .       . 

Ex.130.  8X4-X-Xr8x4x7)<25,    but   by  the   ax10m  of 

fractions,  this  simplifies  by  striking  out  like  factors  in  numerator 

and    denominator,    and    becomes   - — - — : — p=tt.     This    result 

2X4X1X5    40 

would  have  been  obtained  by  striking    out  numerator  factors 

with  equal  denominator  factors  in  the  different  fractions: 


3       2 
4     2  5     = 


In  multiplication,   the  multiplication  symbol   is  often 
replaced  by  the  equivalents  "  of  "  and  "  times."    Thus: 


•   15     1  ..        5     1    ,5      5 

2X8==2timeS8=2°f8=16 


Observation.     To  multiply   a   quantity   by   a  number 
equivalent  to  dividing  the  quantity  by  its  reciprocal. 

Therefore 


this  may  be  written 


8 
2  = 

8A2' 

5 

8 
6  : 

_5V3_5 
8A6     16' 

Observation.  To  simplify  a  compound  fraction,  i.e.,  the 
ratio  of  one  fraction  to  another  fraction,  multiply  the  numer- 
ator fraction  by  the  denominator  fraction  inverted. 


82  PRACTICAL  MATHEMATICS 

Ex.  131.    Simplify  the  following: 


I 

3 

5k 

6 

3/b 

,*    5 

(a)  -+2k+~- 

+ 

^3 

~2p~ 

T' 

«£ 

6 

,7     ,1 

3 

48 

r> 

WgOf- 

times  — 
lb 

X42' 

8 

>»    3 

3 

W      o> 

w  f» 

2 
1 

5 

2 

6 

</>f, 

2 

9 

Ex.  132.    State  why  the  following  forms  are  not  equivalent. 


/  %  3+z     3  r.   6-*^i 

(a)  5*^5'  (6)   5+**1' 

3+2*      3+z  (/)  ^_±^^! 

5+5x     1+x  (9)   ab*+ab+a*b5*3a2b*, 


W  tt?* 


5+3      1+3' 


(A)  7a6+6cM13ac6, 


(d)  SoT**  w  16«2+Tx^4a?6- 


22.  Laws  of  Numbers.  A  polynomial  is  an  expression 
of  several  algebraic  terms,  such  as  a—b  —  c  —  d.  When 
the  polynomial  consists  of  two  terms  only,  such  as  a+b  or 
a— b,  it  is  called  a  binomial,  and  when  it  consists  of  three 
terms,  such  as  a— b  —  c,  it  is  called  a  trinomial. 


FUNDAMENTAL  OPERATIONS  83 

Ex.  133.  Multiply  (a  —  b)  by  (a  —  b),  i.e.,  square  the  binomial 
(a-b).  a  and  b  are  the  symbols  for  any  two  distinct  quantities 
or  numbers. 

a+b 

a2  the  product  of  a  times  a, 

-\-ab         the  product  of  a  times  b, 

+ab          the  product  of  b  times  a, 

+62  the  product  of  b  times  6, 

The  product  =  a2  +2ab+b2  =  the  sum  of  the  partial 

products. 

In  multiplication,  every  term  of  the  minuend  is  multi- 
plied by  every  term  of  the  multiplier.  The  product  of 
terms  contains  the  factors  of  the  terms,  with  the  exponents 
of  like  letters  added,  (a +b)  is  treated  like  a  single  letter 
when  applying  the  principle  of  adding  exponents'. 

/.  (a+b)  times  (a+b)  =  (a+b)2  =  a2+2ab+b2  =  a2+b2+2ab. 

Observation.  The  square  of  the  sum  of  two  quantities  is 
a  trinomial  and  equals  the  sum  of  the  squares  of  the  two 
quantities,  plus  twice  the  product  of  the  quantities. 

Ex.  134.  Write  the  values  of  the  following  without  per- 
forming the  multiplications : 

(a)  (x+y)2,  (d)  (2a +o)2, 

(b)  (2+y)2,  (e)  (5+20)>, 

(c)  (a +5)*,  (/)  (20+1)'. 

Ex.  135.     Multiply  (a—b)  by  (a-b),  i.e.,  square  the  binomial 
(a—b).     a  and  b  may  represent  any  two  quantities  or  numbers. 
a—b 


a—b 


the  product  of  a  times  a, 
-ab  the  product  of  a  times  ( —  b)i 

—ab  the  product  of  (—6)  times  a 

+b2         the  product  of  b  times  b. 


The  product  =a2—2ab+b2  =the  sum  of  the  partial  products. 

/.  (a-b)  times  (a-b)  =(a-b)2=a2-2ab+b2=a2+b2-2ab. 


84  PEACTICAL  MATHEMATICS 

Observation.  The  square  of  the  difference  between  two 
quantities  is  a  trinomial,  and  equals  the  sum  of  the  squares 
of  the  two  quantities,  minus  twice  the  product  of  the  two 
quantities. 

Ex.  136.  Write  the  values  of  the  following  without  multiplying : 

(a)  (x-y)2,  (d)  (2a  -5)*, 

(b)  (2-y)2,  (e)   (5-20)2, 
to    (a-5)2,                                      (/)    (20-1)2. 

In  (e)  (5-20)2=25-200+400=225,  but  (5-20)2  =  (-15)2. 

Observation.  The  product  of  two  negative  terms  or  the 
square  of  a  negative  term  is  positive.  The  product  of  a  positive 
and  a  negative  term  is  negative. 

Ex.  137.    Multiply  (a+b)  by  (c+d). 

a+b 

c+d 

ac  the  product  of  c  times  a, 

+cb  the  product  of  c  times  b, 

+ad  the  product  of  d  times  a. 

+bd  the  product  of  b  times  d, 


The  product  =  ac  +cb  +  ad  -\-bd=  the  sum  of  the  partial  products. 

Ex.  138.    Multiply  (a+b)  by  (a-b). 

a—b 

a+b 


a2  the  product  of  a  times  a, 

—ab    the  product  of  a  times  (—b), 
+ab    the  product  of  b  times  a, 
—b2  the  product  of  b  times  (—b), 
The  product  =a2—b2  =the  sum  of  the  partial  products. 

Observation.  The  product  of  the  sum  of  two  quantities 
times  the  difference  of  the  same  quantities,  equals  the  square 
of  the  first  quantity,  minus  the  square  of  the  second  quantity. 


FUNDAMENTAL  OPERATIONS  85 

Ex.  139.    Write  the  values  of  the  following  without  multiplying: 

(a)  (a+k)(a-k),  (e)    (5d+3r)(5d-3r), 

(b)  (a+k){b-k),  (/)    (5d+3r)(7/b-3c), 

(c)  (5+d)(5-d),  fe)   (16+10)(16-10), 

(d)  (7-2)(7+2),  (/*)  (10+3)(17-4). 
Ex.  140.     Multiply  (a2  -ab  +b)  times  (a  +6)  =a3  +53. 

a2-ab+b2 
a  +b 


a3-a26+a62  the  product  of  a  times  the  mul- 

tiplicand, 
+a2b  —  ab2+b3    the  product  of  b  times  the  mul- 
tiplicand, 


The  product  =a? +b3  =the  sum  of  the  partial  products. 

Ex.  141.    Multiply  (a2+ab+b2)  times  (a  -b)  =a3-b3. 

a2+ab+b2 
a  —b 


a3+a2b+ab2  the  product  of  a  times  the  mul- 

tiplicand, 
—a2b—ab2—b3    the  product    of   (-6)  times  the 
multiplicand, 


The  product  =a3  —b3  =the  sum  of  the  partial  products. 

Ex.  142.     Divide  (a2+2ab+b2)  by  (a +6). 

In  division,  set  down  dividend  and  divisor  in  descending, 
i.e.,  decreasing  powers  of  one  of  the  letters.  The  first 
term  of  the  quotient  is  obtained  by  dividing  the  first  term 
of  the  divisor  into  the  first  term  of  the  dividend.  The 
quotient  will  contain  the  same  letters  as  are  contained  in 
both  divisor  and  dividend,  with  exponents  of  the  former 
subtracted  from  the  exponents  of  like  letters  of  the  latter. 
The  quotient  times  the  divisor  gives  the  first  subtrahend. 


86  PRACTICAL  MATHEMATICS 

Subtract  the  subtrahend  from  the  dividend,  and  set  down 
the  remainder.  The  remainder  with  the  remaining  terms  of 
the  dividend  constitutes  a  new  dividend,  and  the  operation 
outlined  above  is  repeated,  giving  the  second  term  of  the 
quotient.  The  product  of  the  new  term  of  the  quotient 
times  the  divisor  is  the  new  subtrahend. 


dividend 
a2+2ab+b2 
a  times  (a+6),  a2+  ab 


a-\-b,  divisor 


a       first  term  of  quotient, 
+6  second  term  of  quotient, 


first  remainder,  ab 

b  times  (a+b),  ab+b2 

no  remainder. 

a2+2ab+tf 
•*'   "     a+b 


=  a+b. 


Ex.  143.    Divide  a3+V  by  (a2-ab+b2) 

a2— ab+b2  divisor, 
a       first  term  of  quotient, 
+6  second  term  of  quotient, 


dividend 
a3  +bs 

first  subtrahend,      a3  -a2b  +ab2 


first  remainder,  a2b—ab2 

second  subtrahend,        a2b—ab2+b3 
no  remainder. 

a*+b3 


a2 -ab+b'' 


=  a+b. 


Ex.  144.    Perform  and  verify  the  following  divisions : 
,      a2-2ab+b2  a2+5a+G 

a2-6!  .  ,_  a2 -a -6 

a—o  a+z 

(c)  — -J--0-6,  to)     „   0    =q+3, 


a+6             '  0-2 

+6a+9  a' -6' 

— =a+6,  (n) — 

a+3  a-b 


04^9  =o+3>  (fc)  a^-M  =a2+o6+6>. 


FUNDAMENTAL  OPERATIONS  87 

23.  Factoring.  Any  product  may  be  resolved  into  its 
constituent  factors,  by  performing  a  division  which  leaves 
no  remainder.  Both  quotient  and  divisor  are  the  factors 
of  the  dividend. 

Often  the  work  is  simplified  without  division,  by  recog- 
nizing the  factors  by  inspection.  The  simplest  case  is  one 
in  which  each  term  possesses  a  factor  common  to  every  other 
term.  The  product  of  the  common  factors  (H.C.F.), 
is  written  outside  a  parenthesis  and  the  remaining  factors 
of  each  term  are  written  within  the  parenthesis. 

Given  Expression  H.C.F.  Remaining  Factors 

3ax2b+9a2^b2+15ax4b3=3ax2b(l+Saxb+5x2b2). 

Observation.  The  highest  common  factor  of  an  expression 
is  the  product  of  the  common  numeric  and  literal  factors. 
The  least  exponent  of  any  letter  in  the  H.C.F.  is  the  least  ex- 
ponent  of  that  letter  found  in  any  term  of  the  expression. 

Ex.  145.     Factor  the  H.C.F.  from  the  following  expressions: 

(a)  5ax2+25x3+75ax6, 

(6)  25c0s2+12.5c203s4+62.5c20s, 

(c)  216a62+864a6+864a. 

Ex.  146.    Factor  6x2  -14x6  -9x6  +2162. 

There  is  no  H.C.F.  for  this  expression,  although  there  is  a 
common  factor  of  x  for  all  the  terms  except  the  last,  and  a 
common  factor  6  for  all  the  terms  except  the  first.  There  is 
also  a  numeric  factor  of  3  for  all  the  terms  except  the  second. 
In  such  cases  it  is  advisable  to  factor  the  expression  in  two  groups 
of  two  terms  each  as  indicated  by  the  parentheses. 

6x2  -14x6  -9x6+2162  =  (6x2  -9x6)  +( -14x6+2162). 

3x  is  the  common  factor  in  the  first  parenthesis,  and  therefore 

(6x2-9x6)=3x(2x-36). 

—76  is  the  common  factor  in  the  second  parenthesis,  and  therefore 

(-14x6+2162)  =  -7b(2x-36). 


88  PEACTICAL  MATHEMATICS 

(—76)  was  factored  instead  of  (+76)  so  that  both  parenthetical 
factors  might  be  equal. 

.*.     Gx*-14xb-9xb+21b*=3x(2x-3b)-7b(2x-3b). 

The  last  expression  is  therefore  a  difference  of  two  products 
each  having  a  common  factor  (2x  —  3b)  and  therefore  we  obtain 

Sx(2x-3b)  -7b(2x-3b)  =(2x-3b)(3z-7&), 

/.     Gx*-Uxb-9xb+21b*  =  (2x-3b)(3x-7b). 

Verify  this  result  by  multiplying  the  factors. 

Ex.  147.  Factor  the  following  expressions  by  grouping  in 
two  different  ways: 

(a)  ac+ad+cb+bd, 

(6)  5g+kg+ks+5s, 

(c)  9sa+21st+6ap+Utp. 

The  factors  of  any  product  may  be  detected  by  inspection 
or  by  division.  A  familiarity  with  the  types  of  factors 
that  produce  binomial  and  trinomial  products  suffices  for 
most  cases. 

If  the  expression  is  the  difference  between  two  squares, 
such  as  x2— y2,  then  the  factors  are  (x+y)  and  (x— y). 
There  are  no  factors  for  the  sum  of  two  squares.     Why? 

If  the  expression  is  the  difference  between  two  cubes, 
such  as  x3— y3,  then  the  factors  are  (x— y)  and  (x2+xy+y2). 

If  the  expression  is  the  sum  of  two  cubes,  such  as  x3+y3, 
then  the  factors  are  (x+y)  and  (x2— xy+y2). 

If  trinomial  expressions  are  to  be  factored,  we  observe 
first  if  they  are  the  squares  of  binomials,  from  either  of  the 
following  forms  and  their  corresponding  factors: 

x2+y2+2xy  =  (x+y)(x+y)  =  (x+y)2, 

x2+y2— 2xy  =  (x— y)(x— y)  =  (x— y)2. 


FUNDAMENTAL  OPERATIONS  89 

The  sign  of  the  product  terms  (-\-2xy)  or   (  —  2xy)  is 
also  the  connecting  sign  inside  the  binomial  parentheses. 

Ex.  148.    Write  the   factors  of   the  following  expression  by 
inspection  : 

(a)  49-1662,  (d)  a262+2a6+l, 

(6)  8-2763,  (e)   z2+4z+4, 

(c)   tey*+$x',  (/)  (a+6)2-(a-6)2. 


If  the  trinomial  is  not  a  perfect  square,  test  it  for  a 
common  factor.  Then  resort  to  the  following  procedure 
in  order  to  put  it  into  a  four-term  polynomial,  after  which 
it  can  be  factored  by  grouping. 

Factor  6x2-23x6+2162. 

There  is  no  common  factor  for  this  expression. 

Multiply  the  coefficients  of  the  square  terms  6X21  =  126. 
Resolve  126  into  two  factors  whose  sum  is  23,  i.e.,  equals 
the  coefficient  of  the  remaining  term  23xb . 

126  =  53X2  =  42X3  =  21X6  =  18X7  =  14X9,  but  14+9  =  23 

therefore  (  —  23s6)  may  be  replaced  by  ( —  9xb  —  14x6) . 
The  example  becomes  6s2  —  14x6  —  9s6  +  2162,  and  therefore 
the  factors  are  (2s -36)  (3s -76).     See  Ex.  146. 

Ex.  149.     Factor  the  following  trinomials.  ■ 
(a)  3z2-17x+10, 
(6)   10a:2-17x+3, 

(c)  6x2y2-23xyb+20b2, 

(d)  6a2+7a-20. 

In  the  last  example  6  times  (—20)  =  — 120,  therefore  one  of 
the  factors  is  positive  and  the  other  negative.  The  excess  of  one 
factor  over  the  other  equals  7,  i.e.,  7a  =  15a—  8a. 


90  PEACTICAL  MATHEMATICS 

24.  The  Oblique  Triangle. 

Ex.  150.  Draw  an  oblique  triangle,  i.e.,  a  triangle  whose 
sides  are  oblique,  so  that  all  of  its  angles  are  acute.  Draw  the 
three  perpendiculars  by  joining  each  vertex  to  its  opposite  side.' 
If  a  perpendicular  falls  outside  the  triangle  produce  the  side  to 
meet  it. 

Draw  an  oblique  triangle  one  of  whose  angles  is  obtuse.  Draw 
the  three  perpendiculars  by  joining  the  vertices  to  the  opposite 
sides. 

Each  of  these  constructions  gives  the  projections  of  two  sides 
of  a  triangle  upon  the  third  side. 

Ex.  151.  Draw  two  unequal  adjacent  right  triangles  ABD 
and  BDC,  with  a  common  equal  altitude  BD.     The  right  triangles 

form  an  oblique  triangle  ABD.    In  triangle  ABC,  the  sin  A  =  — , 

An 

7?D 
and  therefore  BD=AB  sin  A.    In  triangle  BDC,  the  sin  C  =  r—, 

BL 

and  therefore  BD  =BC  sin  C.    Since  BD  is  an  identity,  i.e.,  a 

common  side  of  both  triangles,  we  say  BD  of  triangle  ABD=BD 

of  triangle  BDC.     Then  by  equality  axiom,  AB  sin  A  =BC  sin  C. 

In  triangle  ABC,  A B  is  opposite  angle  A,  and  may  be  abbreviated 

by  a,  and  BC  is  opposite  angle  C  and  may  be  abbreviated  by 

c.    Therefore 

csin  A  =asin  C. 


sin  A     sin  C 
by  equal  products  theorem  in  proportion. 

Construct  perpendiculars  from  A,  and  also  from  C,  to 
the  opposite  sides  of  triangle  ABC,  and  establish  the  com- 
plete fact 

a  c  b 

sin  A     sin  C     sin  B 

Observation.  In  any  triangle  the  ratio  of  any  side  to  the 
sine  of  its  opposite  angle  is  a  constant.  This  is  known  as 
as  the  Law  of  Sines. 


FUNDAMENTAL  OPERATIONS  91 

If  any  three  parts  of  a  triangle  are  given,  including 
two  sides  and  an  angle  opposite  one  of  them,  or  two  angles 
and  a  side  opposite  one  of  them,  then  this  formula  may  be 
used  to  solve  the  triangle. 

Ex.  152.    Given  a  =  6,  b  =  3,  and  A  =30°. 

To  determine  B,  we  use  — — -  =— — -,  from  which  we  obtain 
sin  A     sin  B 

sin  B= ,    and   substituting   the   above   values   we  obtain 

a 

_     3Xsin30°     3X.5      oc 

sin  B  = =  —  —  =  .25. 

6  o 

In  Table  VIII  locate  .2500  under  the  heading  sines,  and  deter- 
mine the  angle  B  =  14°  30'. 

C  =  180°  -  (30°  + 14°  30Q  =  145°  30'. 
Solve  for  c. 

Ex.  153.     Construct  an    oblique  triangle   ABC,  draw  a  per- 
pendicular h  from  the  vertex  B  to  the  opposite  side  b.     The  side 
b  will  be  divided  into  two  segments  x  and  y,  which  are  the 
respective  projections  of  c  and  a,  upon  b. 
Then 

c2=h2+x2, 

from  the  law  of  the  right  triangle, 

and  x=b—y, 

because  the  sum  of  the  parts  equals  the  whole, 

.*.  x2  =  (b-y)2=b2+y2-2by, 

by  squaring  both  members  of  the  preceding  equation.  This 
value  of  x2  is  now  substituted  in  the  first  equation. 

/.  c2=h2+y2+b2-2by; 

in  the  latter  we  replace  h2+y2  by  a2  because  h2+y2=a2,  also 
replace  y  in  2by,  by  its  equal,  a  cos  C,  because  y=a  cos  C  from 
the  definition  of  a  cosine.  These  substitutions  will  give  the  final 
form 

c2=a2+b2-2ab  cos  C. 


92  PRACTICAL  MATHEMATICS 

Observation.  The  square  of  any  side  of  a  triangle  equals 
the  sum  of  the  squares  of  the  other  two  sides,  minus  twice  the 
product  of  the  two  sides  times  the  cosine  of  the  angle  included 
by  them. 

This  is  known  as  the  Law  of  Cosines. 

a2=b2-fc2  — 2bc  cos  A    and    b2  =  a2+c2— 2ac  cos  B 

are  two  other  forms  for  the  law  of  cosines. 

Ex.  154.  Solve  the  triangle  ABC,  given  a  =  6,  6  =  10,  C=30°. 
The  working  formula  is  c2  =a2+62  -2ab  cos  30°.  Substituting  the 
values  of  a,  b,  and  C  from  the  data  we  obtain: 

c2=36+100-2x6xl0X.866  =  136-104=32, 

c=5.66. 


The  angles  A  and  B  may  then  be  obtained  by  the  law  of  sines. 

a  c       „  ,  .  ,  ....     a  sin  C 

j,  from  which  we  obtain  sin  A  =  - 


sin  A     sin  C  c 

b           c       .           .  .  .           .  t   .      .     n    b  sin  C 
from  which  we  obtain  sin  B  = . 


sin  B    sin  C 

Solve  for  A  and  B. 

Ex.155,  a  =8 
6=6 
c=5.5 

A=? 
£  =  ? 
(?  =  ? 

The  working  formulas  are: 
fl2  =b2+c2—2bc  cos  A,  from  which  we  obtain  cos  A  = — . 

a2_|_C2_^2 

62=a2+c2—  2accosB,  from  which  we  obtain  cos  5= . 

'  2ac 

(7  =  180°  -(A  +B). 

Substitute  the  numeric  values  from  the  data  and  check  by 
graphic  solution. 


FUNDAMENTAL  OPERATIONS  93 

Ex.  156.  The  triangle  in  Ex.  155  may  be  solved  by  con- 
structing the  perpendicular  h,  from  A  to  the  opposite  side  a. 
The  side  a  will  be  divided  into  two  segments  x  and  y,  which  are 
the  projections  of  b  and  c  respectively.  Then  x+y=a.  By  the 
law  of  right  triangles  b2  -x2  =h2,  and  c2  -y2  =h2. 

.'.     b2-x2=c2-y2,      or      b2-c2=x2-y2. 

Observation.  The  difference  of  the  squares  of  two  sides 
of  a  triangle  is  equal  to  the  difference  of  the  squares  of  their 
respective  projections  on  the  third  side.  This  is  known  as 
the  Law  of  Projective  Segments. 

From    the    above    we    may    also    write    (b-\-c)(b  —  c)  = 

(x+y)(z-y),  by  factoring,  or  (x-y)= — /^      cl  =  —— . 

{x+y)  a 

The   last   formula   enables   us   to   compute  the   difference 

between  the  segments.     By  combining  the  sum,  with  the 

difference  of  the  segments,  either  by  addition  or  subtraction, 

we  obtain  either  twice  x  or  twice  y  respectively. 

Solving  for  (x—y)  by  substituting  the  data  from  Ex. 

155,  we  obtain 

b2-c2    36-30.25     _0     , 

.718,     but    x+y  =  a  =  8, 


x+  y  =  S 
by  subtraction  x—  y—    .718 


*     y  — 

a 

8 

—  .i 

• 

x+y  = 

=  8 

by  addition 

x- 

-y= 

=    .718 

b} 

2x 

=8.718 

X 

=4.359 

cos 

B  = 

=y_z 

c 

3.641 
5.5 

.662. 

cos 

\C- 

X 

~b~~ 

4.359 
5.5 

.792. 

2?/  =  7.282 
2/  =  3.641 


5  =  48.6°. 


C  =  37.7°. 


A  =  180o-(A+£)  =  180o-86.3o  =  93.7°. 


94  PRACTICAL  MATHEMATICS 

Ex.  157.     Solve  the  following  triangles  given : 


(a)  a=  5, 

6=6, 

A  =45°, 

(6)  a=  3, 

6=3.2, 

C  =  120°, 

(c)   a  =  12, 

6  =  11, 

C=40°, 

(d)  6  =  11, 

c=5.3, 

4=55.8°, 

(e)  a  =  13, 

6  =  12, 

c=5, 

(/)    6  =  .10,       c  =  .25,         (7  =  105°. 

25.  Logarithms.  We  have  seen  how  one  number  may 
result  from  uniting  other  numbers  by  addition,  subtraction, 
multiplication,  or  division.  A  number  may  be  obtained  as 
a  result  of  a  root  or  a  power  of  another  number.  Thus 
64  may  result  from  adding  50  and  14  or  by  subtracting 
36  from  100  or  by  multiplying  16  times  4  or  dividing  320 
by  5.  Again  64  =  V4096,  and  further  64  =  82  =  43  =  26.  In 
like  manner  729  may  be  considered  the  square  root  of 
53'14'41,  or  729  =  36  =  93  =  272. 

By  the  law  of  multiplication,  we  add  exponents  of 
like  factors,  and  write  this  sum  over  the  like  factor  in  the 
product.  Therefore  321  X358X3*  =38 4.  If  one  or  more  of 
the  exponents  should  be  negative,  the  sum  of  the  exponents 
will  bear  the  sign  of  the  excess.  Thus  1032X10~26X102-4 
=  103  =  1000,  since  3.2-2.6+2.4  =  3,  and  103  =  1000. 
1000  is  the  product  of  three  numbers  a,  b,  and  c,  correspond- 
ing respectively  to  103'2,  10-26,  and  1024,  in  other  words, 
abc=  1000,  where  a  =  1032,  6  =  10"26,  and  c=1024. 

Observation.  An  entire  multiplication  may  be  made  by 
considering  the  factors  and  the  product  as  powers  of  10. 

Our  familiarity  with  powers  and  roots  of  10  is  limited 
as  follows: 

The  first  power  of  10      =  101  =  10. 
The  second  power  of  10  =  102  =  100. 
The  third  power  of  10    =  103  =  1000. 
The  fourth  power  of  10  =  104  =  10000. 
The  fifth  power  of  10     =  105  =  100000. 


FUNDAMENTAL  OPERATIONS  95 

The  square  root  of  10  =  VH)  =  3.162  approx.  =  10*  =  10'6, 
the  one-half  power  of  10. 

The  cube  root  of  10  =  ^10  =  2.154  approx.  =  10*  =  10  333, 
the  one-third  power  of  10. 

The  fourth  root  of  10  =  v'lO  =  1 .778  approx.  =  10*  =  10  25, 
the  one-fourth  power  of  10. 

V10XV10  =  10  5X105  =  10^X10*  =  10 

^10X^/10X^10  =  10*X10*X10*  =  10. 

...  10  X105  =1015     =     10X3.162  =  31.62. 

The  logarithm  of  31.62  =  1.5. 

102X^/lO  =  102-333=   100X2.154=215.4. 

The  logarithm  of  215.4  =  2.333. 

103Xv/10  =  10325  =1000X1.778  =  1778. 

The  logarithm  of  1778  =  3.25. 

From  the  above  it  will  be  seen  that  all  numbers  may 
be  considered  as  powers  of  10.  Such  a  foundation  number 
as  10,  to  which  all  numbers  are  referred  as  powers,  is  called 
a  base,  and  the  exponent  of  the  base  takes  the  special 
name  of  logarithm. 

Observation.  To  obtain  the  logarithm  of  a  number  means, 
therefore,  to  determine  the  corresponding  exponent  of  the 
base  10. 

The  power  of  10  and  the  given  number  are  equivalent, 
i.e.,  103  25  =  1778,  and  therefore  the  logarithm  of  1778  is  3.25. 

Considering  the  integral  powers  of  10  in  the  above  table, 
we  observe  that  the  logarithms  increase  by  a  constant 
difference  of  1,  for  every  higher  multiple  of  10.  There 
are  as  many  zeros  following  1  in  the  given  number  as 
the  corresponding  exponent  of  10.  In  other  words,  a  six- 
figure  number  has  a  logarithm  of  5;  a  five-figure  number 
has  a  logarithm  of  4 ;  a  four-figure  number  has  a  logarithm 
of  3 ;   a  three-figure  number  has  a  logarithm  of  2 ;    a  two- 


96  PRACTICAL  MATHEMATICS 

figure  number  has  a  logarithm  of  1.  In  each  case  the  loga- 
rithm is  one  less  than  the  number  of  integral  figures.  If 
we  divide  a  number  by  10,  we  decrease  its  integral  figure  by 
1,  and  therefore  decrease  its  logarithm  by  1. 

...     102    =ioo.  ®=ai-i  =  ao   but  -=1. 

a  a 

101     =   10.  >.     a°  =  l. 

lOi-i    =10°    =     1.     =^  =  j^.      •••    10°  =  1. 
100-1    =10-i=     0.1 

io-i-i  =  io-2=    0.01 

Observation.  The  zero  power  of  10  or  any  base  equals 
unity.    Logarithms  of  decimals  are  negative. 

A  number  of  two  figures,  such  as  27,  will  have  a  log- 
arithm greater  than  1  but  less  than  2,  because  27  is  greater 
than  10,  but  less  than  100.  A  number  of  three  figures, 
such  as  546,  will  have  a  logarithm  greater  than  2  but  less 
than  3.  The  logarithm  of  a  digit  is  a  decimal  number, 
i.e.,  it  is  greater  than  0  but  less  than  1. 

The  logarithm  of  any  number,  except  a  multiple  of  10, 
consists  of  an  integral  part,  called  the  characteristic,  and 
a  decimal  part,  called  the  mantissa.  The  characteristic  is 
a  positive  integer  for  all  numbers  greater  than  unity,  and 
is  numerically  one  less  in  unit  value  than  the  number  of 
integral  figures,  i.e.,  figures  written  to  the  left  of  the  decimal 
point.  The  characteristic  is  negative  for  any  number 
less  than  one.  Its  unit  value  is  numerically  the  same  as 
the  decimal  position  of  the  first  significant  figure,  i.e., 
first  digit  to  the  right  of  the  decimal  point.  Mantissas 
are  read  in  a  table  of  logarithms  such  as  Table  IX.  Any 
multiple  of  10  has  a  zero  mantissa. 

The  position  of  the  decimal  point  in  a  given  number 
affects   the   unit   value   of   the   characteristic   only.     The 


FUNDAMENTAL  OPERATIONS  97 

mantissa  depends  upon  the  order,  i.e.,  succession  or  sequence, 
of  figures  only.  Thus  the  log  1778  =  3.25.  Dividing  1778 
by  10  gives  177.8,  and  therefore  log  177.8  =  2.25;  again 
dividing  177.8  by  10  gives  17.78  and  therefore  log  17.78  =  1 .25. 
.'.  log  .1778  =  1.25.  The  minus  sign  is  put  over  the  char- 
acteristic because  the  mantissa  is  always  positive. 

Table  IX  consists  of  twenty  vertical  columns.  Column 
one  is  headed  N,  columns  2  to  11  are  headed  "  the  third 
figure  of  your  number,"  and  columns  12  to  20  inclusive 
are  headed  proportional  parts.  The  decimal  points  are 
omitted  from  the  table,  and  should  be  prefixed  to  the 
four  figures  of  the  mantissa.  Table  IX  is  a  four-place 
table.  Other  tables  may  give  3,  5,  6,  7,  10,  or  15  figures 
for  the  mantissa.  The  degree  of  accuracy  required  in  a 
given  computation  determines  the  proper  number-place 
table  which  should  be  used.  For  practical  purposes  a 
four-place  table  suffices. 

To  use  Table  IX,  locate  the  first  two  figures  of  the 
given  number  under  the  column  headed  N,  carry  the  finger 
horizontally  across  the  page,  and  in  the  column  headed  by 
the  third  figure  read  the  mantissa,  i.e.,  the  four  figures. 
Precede  these  figures  by  a  decimal  point,  and  the  latter 
by  the  proper  characteristic  determined  by  inspection. 
Thus  to  find  the  log  of  325,  locate  32  under  the  N  column, 
and  in  the  seventh  column  headed  5  read  5119.  The 
characteristic  equals  2.  .*.  log  325  =  2.5119  .  To  logarize 
a  number  means  to  determine  its  logarithm. 

Ex.  158.     Logarize  and  verify  the  following: 


(a)  log  320=2.4771, 
(6)  log  408=2.6107, 

(c)  log  912=2.9600, 

(d)  log  5.55  =0.7443, 

(e)  logx      =0.4971, 


log  3.2  =1.4771, 
log  4080  =3.6107, 
log  0.912  =1.9600, 
log  0.0555  =2.7443, 

log^        =1.4971. 


98  PRACTICAL  MATHEMATICS 

The  mantissa  of  a  two-figure  number,  such  as  12,  is 
the  same  as  the  mantissa  of  the  number  120.  Therefore 
the  mantissa  of  a  two-figure  number  is  read  in  the  second 
or  zero  column,  opposite  the  two  figures  in  the  N  column. 
The  mantissa  of  a  one-figure  number,  such  as  7,  is  the 
same  as  the  mantissa  of  700.  Any  number  of  zeros  pre- 
ceding or  following  a  given  number  does  not  change  its 
mantissa  (man). 

man  7  =  man  70  =  man  700  =  man  7000  =  man  .0007,  etc. 

To  read  the  mantissa  of  a  four-figure  number,  such  as 
238.4,  we  must  appreciate  the  fact  that  the  number  238.4 
lies  between  238  and  239.  In  other  words,  the  man  238 
is  smaller  than  the  man  238.4,  and  the  man  239  is  greater 
than  the  man  238.4.  The  required  man  238.4  may  be 
obtained  by  interpolation,  i.e.,  adding  a  correction  to  man 
238  or  subtracting  a  correction  from  man  239. 

man  239  =  3784 
man  238  =  3766 


difference  =     18 

An  increase  of  1  in  the  numbers  corresponds  to  an  increase 
of  18  in  the  mantissas.     Therefore  an  increase  of: 

.1  in  numbers  =  18 X.l  =  2  appproximately  to  be  added 
to  the  man  238,  and  an  increase  of  .2  in  numbers  =  18 X-2  =  4 
approximately  to  be  added  to  the  man  238,  and  an  increase 
of  .4  in  numbers  =  18 X.4  =  7  approximately  to  be  added 
to  the  man  238. 

=  means  corresponds  to. 
/.  man  238.4=man  238+ correction  for  .4  =  3766+7 

=  3773. 
.*.  man  238.4  =  man  239  -  correction  for  .6  =  3784-11 

=  3773. 


FUNDAMENTAL  OPERATIONS  99 

Instead  of  calculating  the  correction  for  the  fourth 
figure  of  a  number,  we  may  read  the  correction  under  the 
columns  12-20  headed  proportional  parts.  After  locating 
the  mantissa  for  the  first  three  figures,  continue  the  finger 
horizontally  across  the  page,  and  in  columns  of  proportional 
parts  headed  by  the  fourth  figure  read  the  correction. 

The  mantissa  of  238  is  3766,  and  under  column  4  of 
proportional  parts  is  7.     .\  man  238.4  =  3766+7  =  3773. 

Ex.  159.     Verify  the  following: 

(a)  log  55.55      =1.7447,  (d)  log  0.1764=1.2465, 

(6)  log   9097      =3.9589,  (e)  log  1.985   =0.2978, 

(c)   log  12330      =4.0910,  (/)   log  700.7    =2.8455. 

Ex.  160.  Multiply  55.55  by  700.7.  Since  55.55  =1017447  and 
700.7  =102-8455,  then 

55.55  x700.7  =  1017447  XlO28455  =  1017447+2-8455  =1045902 

= product. 
.'.     man  product  =  .5902. 

Table  IX  serves  a  new  purpose,  viz.,  to  find  the  number 
when  the  mantissa  is  known.  The  process  is  the  reverse  of  finding 
a  logarithm  and  is  called  antilogarization.  The  number  is  the 
antilogarithm  (log-1  or  antilog)  of  a  logarithm. 

Locate  5902  under  columns  2  to  11  inclusive,  and  on  the  same 
horizontal  line,  read  the  antilogarithm  under  the  N  column. 
When  the  mantissa  is  not  given  in  the  table,  read  that  anti- 
logarithm  in  the  N  column  which  corresponds  to  the  next  lowest 
mantissa,  and  correct  the  antilogarithm  by  an  accretion  taken 
from  the  proportional  parts  column.  5899  is  the  next  lowest 
mantissa  to  5902,  and  this  corresponds  to  the  antilogarithm  389. 
The  difference  between  5902  and  5899  is  3,  which  is  located  under 
column  headed  3  of  proportional  parts.  Therefore  the  fourth 
figure  of  the  antilogarithm  of  5902  is  3,  and  accordingly  the 
product  is   38930.    Five  integral  figures  appear  to  the  left  of 

the  decimal  point  because  the  characteristic  was  given  as  4. 
/.     log-1  4.5902  =38930  =product  55.55  X700.7. 

Verify  this  result  by  actual  multiplication. 


100  PRACTICAL  MATHEMATICS 

Observation.     The  logarithm  of  a  product  equals  the  sum 
of  the  logarithms  of  its  factors. 

Ex.  161.    Perform  the  following  multiplications  by  logarithms: 
(a)  38.92X6005,  (d)  72x72x72, 

(6)  7891X.8032,  (e)   100x62.54, 

(c)  360X360,  if)   .0001X83620. 

Ex.  162.     Divide  55.55  by  700.7.    This  may  be  written: 

55>55  _  1Ql"  ^    _  i  nl-7447  -2.8455 

700.7     102-8445 

by  the  law  of  division. 

1.7447  -2.8445  =  1.7447  -  .8455  -2  =  .8992  -2  =2.8992. 

The  subtraction  must  be  performed    so  as  to  leave  a  positive 
mantissa. 

Quotient  =  102-8992. 

Looking  up  the  antilog  of  2.8992  we  obtain  7929. 

The  negative  characteristic  indicates  that  the  first  significant 
figure  is  in  the  second  decimal  place. 

Therefore 

log-1  2.8992  =  .07929=^ 


Verify  this  result  by  actual  division. 
Ex.  163. 

<  x   38-92  t*       1 

(tt)6005'  (rf):6254' 

(KS     *§?!  (  \    _L_ 

W  .8032'  {e)  6.254' 

W    9'  {J)  83620' 

Observation.     The  logarithm  of  a  quotient  equals  the  log- 
arithm of  the  numerator  minus  the  logarithm  of  the  denominator. 


FUNDAMENTAL  OPERATIONS  101 

In  example  161  (c),  360X360  may  have  been  written 
3602.  The  logarithm  (5.1136)  of  the  power  3602  is  twice 
the  logarithm  (2.5563)  of  the  number  360. 

In  Ex.  161  (d),  72X72X72  may  have  been  written  723. 
The  logarithm  (5.5719)  of  723  is  three  times  the  logarithm 
(1.8573)  of  the  number  72. 

Observation.  The  logarithm  of  a  power  of  a  number  is 
equal  to  the  exponent  times  the  logarithm  of  the  number. 

Ex.  164.  Perform  the  following: 
(a)  53601-6,  (d)  360*, 

(6)  8.52-5,  (e)  560-2, 

(c)  92.1-5,  (/)  .018-2. 

Ex.   165.     (c),   Ex.   164,   may  be  written  92.1*  =  V92l  and 

since  log  92.1*  4  log  92.1  J**!,  then  log V92J  J^A. 

Ex.  164  (d)  may  be  written  360*  »  4360,  and  since 
log  360*4  l0g  360  =  !2Sir^  then  l0S  ^360  =  1°?|52. 

Observation.  The  logarithm  of  the  root  of  a  number  equals 
the  logarithm  of  the  number  divided  by  the  root  index,  i.e., 
root  numeral.  A  root  of  a  number  is  equivalent  to  a  power 
of  the  same  number  whose  exponent  is  the  reciprocal  of  the 
root  index. 

Ex.  166.    Perform  the  following: 

i  .    V  360  /  7.    .5/^- 

(a)   T7==,  (d)  V3, 

^92.1 


.01/ 


(b)   5V389,  (e)  Vl.009, 

1.0137 

vToi" 


2.5 


(e)  V2.5  (/) 


In  the  above  outline  on  logarithms,   all  numbers  are 
considered  as  powers  of  the  base  10.     This  system  is  most 


102  PRACTICAL  MATHEMATICS 

commonly  used,  and  the  ordinary  logarithm  table  is  con- 
structed upon  this  so-called  common  base.  It  is  less 
frequently  designated  as  the  Briggs  system.  Its  principal 
advantage  lies  in  its  decimal  characteristic. 

There  are  circumstances  in  limited  branches  of  science 
where  the  bases  2,  2.4,  2.5,  and  other  numbers  are  used 
instead  of  the  base  10. 

Natural  phenomena  are  very  often  formulated  in  terms 
of  an  unending  number  2.71828+ ,  which  is  abbreviated 
by  the  Greek  letter  e,  and  sometimes  by  the  English  e. 
It  becomes  desirable,  therefore,  to  express  numbers  as 
powers  of  s,  and  hence  we  have  tables  of  logarithms 
based  upon  e  instead  of  10.  These  are  called  by  various 
names  as  either  natural  or  Napierian  or  hyperbolic  logarithms. 

The  Napierian  tables  are  used  in  the  same  manner  as 
the  Briggs  tables.  The  former  are  distinguished  by  the 
fact  that  they  give  both  the  characteristic  as  well  as  the 
mantissa  of  a  number,  and  comparatively  larger  numeric 
values  than  the  corresponding  Briggs  logarithm.  The 
Napierian  logarithm  is  abbreviated  loge  in  order  to  dis- 
tinguish it  from  the  common  log  or  logio. 

log  1000  =  3  by  the  Briggs  system, 
loge  1000  =  6.9078  by  the  Napierian  system. 

In  order  to  convert  the  Briggs  logarithm  of  a  number 
into  a  Napierian  logarithm  the  former  must  be  multiplied 
by  a  conversion  factor  M  called  a  modulus.  In  other 
words,  loge  1000 =M  log  1000.     Therefore, 

,,    loge  1000     6.9078     OQno,     OQAQ 

M=ioflooo=^-=2-3026=2-303  approx' 

log  10  =  1        loge  10  =  2.303. 
loge  10  =  M  log  10,  /.  log  10  =jj  loge  10  =  .434  loge  10. 

The  reciprocal  of  M=-^  =  ^        =.434. 


FUNDAMENTAL  OPEKATIONS  103 

Observation.  The  smaller  base  produces  the  larger  loga- 
rithm. 

Ex.  167.  Write  the  Napierian  logarithms  of  the  following 
by  multiplying  the  modulus  into  the  common  logarithm,  check 
the  results  by  consulting  a  Napierian  logarithm  table : 

(a)  120,  (d)  .125, 

(6)  32.5,  (e)   .025, 

(c)   1.125,  (/)  log  325. 

Ex.  168.  In  computing  the  logarithm  of  a  power  it  is  desirable 
at  times  to  use  the  logarithmic  operation  twice  in  succession. 
Thus B  =  6820L63  is  solved  as  follows: 

log  B  =  1.63  log  6820 

by  law  of  log  of  a  power. 

;.    log  (log  B)  =log  1.63  +log  log  6820 

by  the  law  of  log  of  a  product. 

log  log  B  is  abbreviated  log"  B  or  LLB 

/.    log  log  6820  =log"  6820  =LL  6820. 

/.     log"  B  -log  1 .63  +log  "  6820. 

log  1.63  =  .2122;  log  6820=3.8338;    log"  6820  =log  3.8338  =  .5836 

.*.     log"  B  =  .2122  +.5836  =  .7958. 

In  order  to  obtain  B  the  operation  of  antilogarization  must  be 
performed  twice 

/.     log  B  =log~1 .7958  =6.248, 

5=  log"1  6.2480, 

5  =  1770000. 


This  principle  may  be  used  in  multiplying  a  common  logarithm 
by  the  modulus  2.303. 


104  PEACTICAL  MATHEMATICS 

Ex.  169.     Obtain  the  Napierian  logarithm  of  .625. 
loge  .625  =2.303  log  .625, 
log  .625  =  1.7959, 
.'.     loge  .625  =2.303  times  1.7959. 

1.7959  is  part  negative  and  part  positive,  so  that  the  product 
will  be  part  negative  and  part  positive.  Therefore  these  two 
parts  must  be  clearly  distinguished  by  one  of  the  following 
methods. 

(I)  1.7959  may  be  united,  giving  the  negative  excess  =  -.2041. 
.*.  2.303 X( -.2041)  =  -.4700.  The  mantissas  of  common  loga- 
rithms are  always  positive  and  are  so  read  in  the  common  log- 
arithm table.  Napierian  tables  may  have  the  decimals  negative 
or  positive.  In  the  latter  case  —.4700  could  not  be  read  in  the 
table.  We  may  overcome  this  difficulty  by  writing  1.5300,  which 
is  equivalent  to  —.4700  and  is  obtained  by  both  adding  and 
subtracting  1  to  —  .4700,  which  becomes : 

-.4700+1-1  =(-.4700+1) -1  =.5300-1  =1.5300, 
/.     2.303  X  (1.7959)  =1.5300, 
/.     loge  .625  =  1.5300. 

(II)  2.303 X (1.7959)  =2.303(  -1  +.7959)  =  -2.303 +  (2.303  X. 7959) 

2.303  X.7959  =  1.833, 
.'.     2.303  X  (1.7959)  =  -2.303+1.833  =  -.4700  =1.5300, 
/.     loge  625  =  -  .4700  =  T.5300. 

Observation.  Logarithms  simplify  arithmetic  processes 
by  reducing  them  to  more  fundamental  operations.  Addition 
and  subtraction  cannot  be  further  simplified.  The  arithmetic 
operations  of  multiplication,  division,  power,  and  root,  are 
changed  by  the  use  of  logarithms  to  the  respective  operations 
of  addition,  subtraction,  multiplication,  and  division. 

26.  The  Slide  Rule.  Write  a  series,  i.e.,  a  row  or  suc- 
cession or  progression  of  numbers  from  0  to  10,  in  gradations, 
i.e.,  by  increases  of  .5.    This  is  abbreviated  by  0,   .5,   1, 


FUNDAMENTAL  OPERATIONS  105 

1.5,  2,  2.5  ...  10;  the  dotted  line  .  .  .  means  and  so  on. 
Directly  under  these  numbers  write  their  respective  squares, 
and  in  the  intervening  spaces  on  the  following  row  (3) ,  write 
the  differences  of  the  square,  and  under  the  latter,  on  row 
(4),  write  the  differences  of  the  latter.     Thus: 

row  1,  1  1.5  2  2.5  3  .  .  .     10 

row  2,  1  2.25  4  6.25  9  ...  100 

row  3,         1.25  1.75      2.25  2.75  .  .  . 

row  4,  .5  . 5 .  .5  . 5  .  .  . 

Rows  1  and  3  are  called  arithmetic  progressions,  i.e., 
series  with  a  constant  difference  between  their  consecutive 
terms. 

Ex.  170.  Determine  the  sum  of  the  arithmetic  progression 
(.5,  1,  1.5,  2,  ,2.5  .  .  .  10).  The  first  term  is  1,  which  we  may 
designate  by  f,  the  last  term  is  10,  which  we  may  designate  by 
1,  the  constant  difference  is  .5,  which  we  may  designate  by  d, 
the  number  of  terms  is  20,  which  we  designate  by  n,  and  finally 
designate  the  sum  by  S. 

term  number    2  =term  number    1  +.5  difference,  =f+d, 
term  number    3  =term  number    2 +.5  difference,  =f+2d, 
term  number    4  =term  number    3 +.5  difference,  =f+3d, 
term  number  20  =term  number  19 +.5  difference,  =f-\-19d=l, 

The  sum  equals  the  average  value  between  first  and  last  term, 
multiplied  by  the  number  of  terms 

Ex.  171.  (a)  What  is  the  sum  of  the  first  ten  numbers?  (b) 
What  is  the  sum  of  the  nine  digits? 

Ex.  172.  Write  a  series  of  digits  from  1  to  10.  Under  these 
write  their  respective  cubes.  On  a  third  row  write  the  differences 
of  the  cubes.  On  the  fourth  row  write  the  differences  of  the 
numbers  of  the  third  row.  What  kind  of  a  series  does  the  fourth 
row  represent? 


106  PRACTICAL  MATHEMATICS 

A  geometric  progression  is  a  series  of  terms  which  preserves 
a  constant  ratio  between  consecutive  terms.  Such  a  series  is 
represented  by  the  powers  of  10,  as  10,   100,   1000,  10000,  .  .  . 

Ex.  173.  Prepare  a  strip  of  cardboard  1  in.  wide  by  10  ins. 
long.  Half  way  between  the  upper  and  the  lower  edges  draw 
a  center  line.  On  the  latter  mark  ten  points  of  division  1  in. 
apart,  labeling  them  0,  1,  2,  .  .  .  10.  Through  these  points 
of  division,  draw  vertical  lines  intersecting  the  upper  and  lower 
edge.  At  the  ten  corresponding  points  of  division  on  the  upper 
edge,  write  the  squares  to  the  numbers  on  the  center  line.  At 
the  corresponding  points  on  the  lower  edge,  write  the  cubes  of 
numbers.  We  now  have  three  scales  which  we  shall  designate 
respectively  as  the  scale  of  squares  (S),  the  scale  of  proportionate 
parts  (PP),  and  the  scale  of  cubes  (C).  Subdivide  the  PP  scale 
into  tenths,  and  supplement  the  S  and  C  scales  by  adding  addi- 
tional graduations  to  correspond  to  the  new  divisions  on  PP. 
The  additional  values  for  S  and  C  may  be  calculated  with  very 
little  trouble,  by  multiplying  the  ten  original  squares  and  cubes 
by  a  proportionate  ratio.  This  rule  serves  the  double  purpose 
of  reading  squares  and  square  roots,  and  also  cubes  and  cube  roots. 
Locate  a  number  on  the  S  scale  and  immediately  opposite  on  the 
PP  is  its  square  root.  Thus  6.25  on  S  is  opposite  2.5  on  PP. 
Locate  a  number  on  the  C  scale  and  immediately  opposite  on  the 
PP  scale  is  its  cube  root.     Thus  64  on  C  is  opposite  4  on  PP. 

Ex.  174.  Prepare  a  strip  of  cardboard  1  in.  by  10  ins.  On 
the  lower  edge  lay  off  a  scale  (L)  of  proportionate  parts  with 
subdivisions  in  tenths  like  the  PP  scale  in  Ex.  173.  On  the  upper 
edge  lay  off  a  scale  (D)  of  logarithms  as  follows:  Opposite  the 
following  points  of  the  L  scale 

0      3.01      4.77      6.02    6.99    7.78    8.45    9.03    9.54     10.00 

place  the  following  respective  graduations  on  the  D  scale : 

12  3  45678910 

If  the  D  scale  is  supplemented  further  by  additional  graduations 
the  rule  may  be  used  in  place  of  a  logarithm  table.  Locate  any 
number  on  the  D  scale  and  immediately  opposite  on  the  L  scale 
is  its  logarithm.  The  extreme  right-hand  reading  of  the  L  scale 
is  considered  1  in  reading  logarithms. 

To  multiply  4  by  2  adjust  the  needle  points  of  a  dividers 
on  the  D  scale  so  that  they  span  the  distance  between 


FUNDAMENTAL  OPERATIONS  107 

1  and  2.  This  span  is  the  logarithm  of  2.  A  span  on  the 
D  scale  between  1  and  4  is  the  logarithm  of  4.  These  two 
spans  can  be  added  by  placing  one  needle  point  of  the 
dividers  at  4,  and  then  with  a  span  equal  to  the  logarithm 
of  2,  the  other  needle  point  will  fall  on  8.  In  other  words, 
the  log  4+ log  2  =  log  8,  and  therefore  4X2  =  8.  The  same 
result  will  be  obtained  by  taking  the  span  of  the  dividers 
between  1  and  4  on  the  D  scale,  and  then  if  one  of  the 
needle  points  is  placed  at  2  on  the  D  scale,  the  other  needle 
point  will  fall  on  8.  Again  we  have  log  2+ log  4  =  log  8, 
or  2X4  =  8.  To  divide  8  by  4,  reverse  the  process  with 
the  dividers,  i.e.,  with  a  span  corresponding  to  log  4,  set 
one  needle  point  at  8,  and  the  other  needle  point  will  fall 
on  2,  showing  log  8  — log  4  =  log  2,  or  84-4  =  2.  Explain 
how  you  would  adjust  and  set  the  dividers  to  divide 
8  by  2. 

The  slide  rule  was  invented  in  order  to  obviate  the  use 
of  the  dividers  as  described  above  and  yet  retain  the 
advantage  of  a  scale  over  a  table.  The  slide  rule  takes 
its  name  from  the  fact  that  it  consists  of  a  grooved  or 
channeled  stick  called  the  staff,  stock,  or  rule,  and  a 
sliding  tongued  strip  of  wood  called  the  slide  or  tongue. 

The  contiguous  lower  edges  of  slide  and  rule  bear  a 
pair  of  like  logarithmic  scales  C  and  D  respectively,  which 
are  identically  the  same  as  the  D  scale  constructed  in 
Ex.  174.  By  pulling  the  slide  out  to  the  right  it  is  possible 
to  add  a  logarithm  on  the  C  scale  to  a  logarithm  on  the 
D  scale.  By  pulling  the  slide  to  the  left  it  is  possible  to 
subtract  a  logarithm  on  the  C  scale  from  a  logarithm  on 
the  D  scale.  The  upper  contiguous  edges  bear  two  like 
logarithmic  scales  A  and  B,  which  are  in  reality  two  C 
or  D  scales  placed  end  to  end  in  the  corresponding  space 
of  C.  The  extreme  left-hand  mark,  the  middle  mark, 
and  the  extreme  right-hand  mark  of  any  scale,  are  known 
respectively  as^  the  left  index,  (lin),  middle  index  (min), 
and  right  index  (rin). 


108 


PEACTICAL  MATHEMATICS 


Additional  scales  shown  in  Fig.  19  are  placed  on  slide 
rules  of  different  makes  and  their  construction  and  use 
is  described  later. 


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Ex.  175.  Multiply  2.5  by  3.5.  Opposite  2.5  on  the  D  scale, 
set  the  (lin)  left  index,  of  C,  and  under  3.5  on  C,  read  8.75  on  D. 
We  have  added  the  log  of  3.5  to  the  log  of  2.5.     One  factor  was 


FUNDAMENTAL  OPERATIONS  109 

located  on  C,  and  the  other  factor  on  D,  while  the  product  is 

read  on  D.  No  attention  is  given  to  the  decimal  point  in  locating 
a  factor,  as  the  scales  represent  mantissas,  and  any  sequence 
of  figures  representing  a  factor  would  be  located  in  the  same  position 
on  the  scale  irrespective  of  its  decimal  value. 

Diagrammatic  Setting.  Proportionate  Setting. 

C  scale        lin  lowbr  edge  of  slide   35  Clin      3.5  1       3.5 

D  scale    25  contiguous  edge  of  rule  875      D2.5=875    °r  2J5  =875 

The  diagrammatic  setting  shows  the  position  of  the  respective 
marks  on  the  C  and  D  scale  for  giving  the  product,  and  is  inter- 
preted as  follows:  set  the  left  index  (lin)  of  C,  over  the  factor 
2.5  on  the  D  scale,  under  the  factor  3.5  on  the  C  scale,  read  the 
product  8.75  on  the  D  scale.     In  other  words  2.5X3.5=8.75. 

The  equation  2.5X3.5=8.75  may  be  written  as  a  proportion, 

1        3  5 

—  =—!—,  which  is  the  form  given  in  the  proportionate  setting. 
2.5     8.75 

If  the  two  numerators  of  the  proportion  are  read  on  the  C  scale, 

then  the  two  denominators  are  read  on  the  D  scale. 

It  will  be  observed  that  every  number  on  the  C  scale  bears  the 
same  constant  ratio  to  the  opposite  number  on  the  D  scale  for  any 
given  setting. 

The  equation  2.5x3.5=8.75  may  be  written  as  proportions 
in  three  other  ways,  each  of  which  corresponds  to  a  different  setting 

C  2.5     8.75    C  lin     2.5     C  3.5     8.75 
of  the  rule  as  follows :  77^—  =  ttt  >  77~<T7  =  TT^T )  7Ti"r_  =  ~ftf- 

Dim     3.5    D  3.5    8.75'  D lin     2.5 

These  forms  follow  from  the  theorem  on  proportions. 

Ex.  176.     Multiply   3.5x6.5.    The   diagrammatic   setting  is 
C  3.5        (rin) 


D  22.75      6.5 
other  forms? 


,  or  any  one  of  its  equivalent  forms.     State  the 


Instead  of  using  the  C  and  D  scales  the  like  operation  may 
be  performed  with  the  use  of  the  A  and  B  scales.  The  latter 
differ  from  the  former,  in  so  far  as  they  represent  two  scales  like 
C  and  D,  placed  end  to  end.  The  graduations  are  therefore 
closer  together,  and  therefore  the  A  and  B  scales  give  less  accurate 
readings  than  the  C  and  D  scales.     The  range  of  values  on  the 


110  PRACTICAL  MATHEMATICS 

A  and  B  scales  is  1  to  100,  whereas  the  range  of  the  C  and  D 
scales  is  1  to  10.    The  settings  for  the  A  and  B  scales  are: 

Diagrammatic  Setting.  Proportionate  Setting. 


Left  half  of  A     3.5 


Left  half  of  B     lin 


22.75  Right  half  of  A        A  3.5    22.75 


6.5    Left  half  of  B  Blin       6.5 


Show  the  other  settings  of  the  A  and  B  scales  for  \he  same  example. 
Ex.  177.  Make  a  complete  outline  of  four  different  settings 
of  the  C  and  D  scales  for  the  following  examples.  State  which 
index  was  us*ed  and  the  corresponding  extensional  direction  of 
the  slide. 

(a)  3.62X8.16,  (e)   4.62x2.08, 

(b)  11.2X19,  (/)  1728X16, 

(c)  9.5X1.1,  (g)  .163X12, 

(d)  1.1X2.1,  (h)  .147 X. 133. 

In  two  of  the  four  settings  the  product  is  read  on  the 
D  scale,  and  they  are  called  direct  settings,  In  the  other 
two  settings  the  products  are  read  on  the  C  scale,  and  they 
are  called  inverse  settings. 

The  position  of  the  decimal  point  of  a  product  may  be 
determined  by  inspection,  otherwise  it  may  be  determined 
by  observing  the  position  of  the  decimal  point  of  each  factor, 
also  the  extensional  direction  of  the  slide,  and  the  direct 
or  inverse  setting.  The  worth  of  a  number  greater  than 
1  is  equal  to  the  number  of  digits  to  the  left  of  its  decimal 
point.  The  worth  of  a  number  less  than  1  is  negative 
and  is  equal  to  the  number  of  zeros  preceding  its  first 
significant  figure. 


Number 

Worth 

3596 

4 

823 

3 

11.2 

2 

9.5 

1 

.16 

0 

.013 

-1 

.0072 

-2 

FUNDAMENTAL  OPERATIONS  111 

For  direct  settings,  the  worth  of  a  product  is  the  sum 
of  the  worths  of  its  factors,  when  the  slide  extends  to  the 
left,  The  worth  of  a  product  is  one  less  than  the  sum 
of  the  worths  of  its  factors,  when  the  slide  extends  to  the 
right. 

For  inverse  settings,  the  worth  of  a  product  is  the  sum 
of  the  worths  Of  its  factors,  when  the  slide  extends  to  the 
right  and  one  less  than  the  sum  of  the  worths  of  its  factors, 
when  the  slide  extends  to  the  left. 

Ex.  178.  Since  the  log  of  2.52  is  twice  the  log  of  2.5,  we  can  use 
scales  A  and  D,  or  B  and  C,  to  obtain  the  square  of  a  number. 
Scales  A  and  B  have  correspondingly  twice  the  range  of  scales 
C  and  B  by  construction.  Therefore  we  locate  the  number  on  the 
D  scale  and  read  its  square  directly  opposite  on  the  A  scale,  or 
we  locate  the  number  on  the  C  scale  and  read  its  square  directly 
opposite  on  the  B  scale.  Opposite  2.5  on  C  or  D  is  6.25  on  B  or 
A  respectively.    6.25  =2.52 

The  slide  rule  is  encircled  by  a  runner,  which  is  a 
sliding  piece  of  glass  mounted  in  a  metal  frame.  The  glass 
is  scratched  with  a  straight  line  and  by  adjusting  the 
scratch,  the  graduations  on  two  scales  may  be  brought 
into  alignment,  i.e.,  into  a  straight  line. 

Ex.  179.     Perform  the  following  indicated  operations: 

(a)  1.45*,  (g)   V«U3, 


(6)  32.22,  (h)   V.3872, 

(c)  6.132,  (i)    V2V27 

(d)  .38722,  (j)   2x22=23=,      • 

(e)  VIM,  (k)  2V2=2*  =  , 

(/)    VSZ2,  (I)   2222  =24  =  . 

« 

Division  is  performed  in  the  reverse  manner  to  multi- 
plication. In  the  use  of  the  slide  rule  the  principle  of 
multiplication  is  to  add  the  logarithms  of  the  factors. 
The  sum  of  the  logarithms  is  the  logarithm  of  the  product. 


112  PEACTICAL  MATHEMATICS 

The  principle  of  division  is  to  subtract  the  logarithm  of 

the   denominator   from   the   logarithm   of  the   numerator. 

The  difference  of  the  logarithms  is  the  logarithm  of  the 

5.7 
quotient.     Thus  to  divide  -7^   subtract  the  log  6.8  from 

0.0 

the  log  5.7.  This  may  be  performed  by  setting  the  divisor 
6.8  on  C,  over  the  dividend  5.7  on  D.  The  slide  extends 
to  the  left,  so  that  instead  of  reading  the  quotient  under 
the  left  index,  we  observe  that  the  right  index  occupies 
the  same  relative  position  from  the  end  of  the  scale,  therefore, 
under  the  right  index  of  C  read  the  quotient,  .838  on  D. 

The  setting  is  7T,r^  =  ~77«j7r,  which  follows  from  proportion 

170.7       .000 

6.8  _    1 

5.7  ".838* 

This  quotient  may  be  located  by  three  other  settings 

(7  5  7      838  C 
obtained   from  the  equivalent  proportions,    jy^~o=:~yjy~> 

6.8 C_  5.7         1C  =.838C 

or   ID      .838'  °r6.8D      5.7  D' 

When  the  quotient  is  read  on  the  D  scale  we  have  a 
direct  setting,  and  when  the  quotient  is  read  on  the  C 
scale  we  have  an  inverse  setting. 

Ex.  180.  Make  a  complete  statement  of  the  settings  of  the 
C  and  D  scales  for  the  following  examples,  stating  which  index 
was  used,  and  also  the  extensional  direction  of  the  slide. 

(a)  3.62-8.16,  (e)  4.62 -.208, 

(6)   112-4-19,  (/)  1728^-16, 

(c)'  9.5 -s- 1.1,  (g)  .163+12, 

(<i)  1.1+2.1,  (h)  .147 +  .133. 

For  a  direct  setting,  the  worth  of  the  quotient  equals 
the  worth  of  the  dividend  minus  the  worth  of  the  divisor 
when  the  slide  extends  to  the  left.  This  difference  is 
increased  by  one  to  give  the  worth  of  a  quotient  when 
the  slide  extends  to  the  right. 


FUNDAMENTAL  OPERATIONS  113 

5  7  1 

The  problem  ^  may  be  written  as  a  product  5.7X^-5, 
b.o  o.o 

and  may  be  solved  as  a  problem  in  multiplication  by  adding 
the  log  5.7  to  the  log  of  the  reciprocal  of  6.8.     CI  is  a 

reciprocal  scale  and  when  used  with  D  the  setting  is   n 
6.8  C 


.838  D' 


The  cube  of  a  number  as  well  as  its  cube  root  may  be 
read  directly  when  the  slide  rule  contains  a  scale  (K)  of 
cubes.  Scale  K  has  three  times  the  range  of  scales  C  and 
D,  i.e.,  it  represents  three  C  scales  placed  end  to  end. 
Since  the  log  3.53  is  three  times  the  log  of  3.5,  we  can  use 
scales  K  and  D  to  find  cubes  of  numbers.  Locate  the 
number  3.5  on  D,  and  by  means  of  the  scratch  on  the 
runner,  we  find  42.88  aligned  on  the  K  scale.  To  find 
^60  locate  60  on  the  K  scale,  and  opposite  on  the  D  scale 
read  3.91. 

When  the  slide  rule  does  not  contain  a  K  scale,  the 

cube  of  a  number  may  be  read  with  the  D,  B,  and  A 

scales.   Since  3.53  =  3.5X3.52, then  log  3.53  =  log3.5+log3.52. 

The  D  scale  gives  the  logs  of  numbers  and  the  B  scale  gives 

the  logs  of  squares  of  numbers.     Therefore  set  the  left 

index  of  C  over  3.5  on  D,  and  over  3.5  on  B,  read  42.88 

C  1      42.88  A      T 
on  A,  or  =  R  .     In  other  words,  by  setting  the 

slide  we  have  added  the  log  3.52  to  the  log  3.5.     The  reverse 

operation  will  give  the  cube  root  of  a  number.     Locate 

27  on  the  A  scale  by  setting  the  runner,  then  pull  out  the 

slide  until  the  reading  3  on  D,  under  the  left  index  of  C, 

is  identical  with  the  reading  3  on  B,  in  alignment  with  the 

CI       27  A 
runner,  or -T-3--. 

The  alignment  of  scales  CI  and  D  sets  all  numbers 
on  the  one  scale  to  their  reciprocals  on  the  other  scale. 
Thus  5  on  D  is  opposite  .2  on  CI;    8  on  CI  is  opposite 


114  PRACTICAL  MATHEMATICS 

.125  on  D.  The  reciprocal  of  an  integral  number  must 
be  read  as  a  decimal. 

The  alignment  of  scale  S  with  A  or  B  enables  us  to 
locate  an  angle  on  the  S  scale,  and  read  its  sine  directly 
opposite  on  the  A  or  B  scale.  Thus  30°  on  S  is  opposite 
its  sine  .5  on  A  or  B. 

The  alignment  of  scale  T  with  C  or  D  enables  us  to 
locate  an  angle  on  the  T  scale,  and  read  its  tangent  directly 
opposite  on  the  C  or  D  scale.  Thus  25°  on  T  is  opposite 
its  tangent  .4663  on  C  or  D. 

The  alignment  of  scales  D  and  L,  enables  us  to  locate 
any  number  on  the  D  scale,  and  read  its  logarithm  directly 
opposite  on  the  L  scale.  Thus  6  on  D  is  opposite  its  log- 
arithm .7782  on  L. 

A  log  log  slide  rule  contains  three  additional  scales, 
LL  1,  LL  2,  LL  3,  which  are  segments  of  one  continuous  scale, 
but  which  are  placed  one  above  the  other  for  convenience. 
The  LL  scale  represents  logarithms  of  logarithms  or  second 
logs,  i.e.,  logs". 

The  alignment  of  the  LL  scales  with  C  or  D,  enables  us 
to  locate  a  number  on  LL  and  read  its  Napierian  logarithm 
on  C  or  D.  Thus  10  on  LL  3,  is  opposite  its  Napierian 
logarithm  2.303  on  C  or  D.  1.25  on  LL  2,  is  opposite  its 
Napierian  log  .2232  on  C  or  D.  1.04  on  LL  1,  is  opposite 
its  Napierian  logarithm  .0392  on  C  or  D. 

The  LL  scales  with  the  C  or  D  scales  may  be  used  to 
find  any  integral  or  decimal  power  of  a  number.  Thus 
to  obtain  41'6,  set  the  left  index  of  C  under  4  on  LL  3,  and 

over  1.6  on  C,  read  the  power  9.19  on  LL  3,  or  — — 

1  on  C 

9.19onLL3  ,+  .     3.24/——        ,     ,_ 

To    obtain    v  1.046,    set    the    runner    to 


1.6  on  C 

1.046  on  LL1,  and  3.24  of  C  under  the  runner,  over  the 
left  index  of  C,  read  the  root  1.01397  on  LL  1,  or 
1.01397  LL1_  1.046  LL  1 
lin  C  SMC     ' 


FUNDAMENTAL  OPERATIONS  115 

The  positions  of  x  and  —  are  usually  indicated  on  a 

slide  rule  because  of  their  constant  use.  Additional  marks, 
called  gauge  points,  are  indicated  for  various  other  special 
computations.  Two  of  these  appear  usually  on  the  S  scale 
as  1',  1",  and  are  known  respectively  as  the  minute  and 
seconds  gauge  points.  They  are  used  in  determining  the 
sines  and  tangents  of  angles  less  than  those  indicated  on 
the  extreme  left  of  the  S  and  T  scales.  Sines  and  tangents 
of  small  angles  are  practically  numerically  equal. 

In  reading  tangents  of  angles  greater  than  45°  and  less 
than  90°,  locate  the  tangent  of  the  complementary  angle 

and  use  its  reciprocal,     tan  60°  =  cot  30°= ■= — —  = 

tan  30°     .575 

1.732.  This  work  may  be  simplified  by  locating  the  com- 
plementary angle  30°  on  T  and  reading  1.732  directly 
opposite  on  CI  scale. 

In  reading  sines  of  angles  between  3.43'  and  34.3',  set, 
the  minute  gauge  point  1'  of  S  under  the  number  of  minutes 
on  A,  and  read  the  sine  on  A,  directly  opposite  the  left 
index  of  S.  This  process  is  applicable  to  smaller  decimals 
of  an  angle,  but  where  the  angle  is  given  in  seconds,  its 
sine  may  be  obtained  by  using  the  seconds  gauge  point 
1"  on  S.  Set  the  gauge  point  1"  of  S  under  the  number 
of  seconds  on  A,  and  read  the  value  of  the  sine  on  A,  over 

the  left  index  of  S.     Thus  for  sin  25'  we  have  ^— — 

lin  on  aS 

25  on  A     ,   .  0  .,    ,    .0001699  on  A     35  on  A 

=  — -;    for  sin  35  we  have  — — =- r-. 

1  on  S  lm  on  S  1  on  S 

/.      sin  25'  =  .00728;  sin  35"  -  .0001699. 


In  reading  tangents  of  angles  less  than  5°  44.9'  convert 
the  angle  measure  into  minutes  or  seconds  according  to 
the  magnitude  of  the  angle.  Use  the  minutes  or  seconds 
gauge  point,  correspondingly  adjusted  to  the  A  scale,  in 


116 


PEACTICAL  MATHEMATICS 


exactly  the  same  manner  as  for  determining  sines  of  angles 
less  than  34.3'  as  described  in  the  preceding  paragraph. 

The  worth  of  the  sine  or  tangent  of  an  angle  is  deter- 
mined from  the  position  of  the  angle  in  Table  VI. 
Sin  90°  =  1.0000;  tan  45°  =  1.0000.  The  sines  of  all  angles 
and  the  tangents  of  angles  less  than  45°  are  decimals. 

TABLE  VI 


Angles. 

Worth  of  sin  or  tan. 

sin  and  tan. 

5°  44.9' 

0 

.1 

34.3' 

-1 

.01 

3.43' 

-2 

.001 

.343' =20.6" 

-3 

.0001 

2.06" 

-4 

. 00001 

Ex.  181.     With  the  use  of  the  runner,  a  continued  product 

may  be  made  quickly  and  accurately.     Thus  multiply  8  Xl2  X7.5. 

lin  C     12  C 
The  product  8  X 12  =96  is  represented  in  the  setting  -— -  =^~^- 

Therefore    8x12x7.5=96x7.5;    the    latter    is    represented    by 

7.5(7     rin(7 

720D~96D* 

This  double  operation  may  be  performed  with  one  setting 

,  ii     -    hnC  \    10  .  7.5  C    .      A 

as  follows:   — —  =  runner  to  12  .  .  .  rm  to  runner  =  ,  in  other 

D  8  7z(J  U 

words,  set  left  index  of  C  over  8  on  D,  move  runner  to  12  on  C, 

slide  right  index  of  C  to  runner,  under  7.5  on  C  read  the  product 

720  on  D.     :.     8x12x7.5=720.  * 


Ex.    182.     Multiply    9x10.5x8x6.3.     Verify    the    following 
setting,  giving  an  analysis  according  to  the  successive  products: 


linC 
Z>9 


= runner  to  10.5  . 


1  to  runner  .  .  . 
runner  to  8  . 


1  to  runner 


6.3(7 
4770  D' 


FUNDAMENTAL  OPERATIONS  117 

in  other  words,  set  the  left  index  of  C  over  9  on  D;  move  the 
runner  to  10.5  on  C;  set  right  index  of  C  to  runner;  move  runner 
to  8  on  C;  set  right  index  of  C  to  runner;  under  6.3  on  C  read 
4770  on  D.     .\     9x10.5x8x6.3=4770. 


9.1  X8.2 
Ex.  183.    Simplify  -1- — -^.    This  may  be  analyzed  by  per- 
7.o  Xo.o 

forming  the  operation  as  follows:    divide  9.1  by  7.3;    multiply 

the  quotient  by  8.2;  divide  the  product  by  3.5.     The  setting  is 

C7.3 

jr-rr-  =runner  to  9.1  ...  1  to  runner  ...  rin  C 

runner  to  8.2  .  .  .  3.5  to  runner  =- 


2.818  D' 


in  other  words,  set  7.3  on  C  over  the  left  index  of  D;  move  runner 
to  9.1  on  C;  set  right  index  of  C  to  runner;  move  runner  to 
8.2;  set  3.5  on  C  to  runner;  under  the  right  index  of  C  read 
2.818  on  D. 

This  operation  may  be  performed  by  using  the  C,  CI,  and 
D  scales,  necessitating  but  one  adjustment  of  the  runner  as  follows: 

8.2  on  CI  .    ..      fm 

91onD  =  TUnmr  to  lm  °f  CI    *  \  9       nt  3 .5  on  CI 

y-1  on  u  7.3  on  C  to  runner =—— 

2.818  on  D 

in  other  words,  set  8.2  on  CI  over  9.1  on  D;  move  runner  to 
left  index  of  CI;    set  7.3  on  C  to  runner;   under  3.5  on  CI  read 

2.818  on  D.     .:     ^^  =2.818. 
7.3X3.5     

Ex.  184.  Multiply  2.4  X6.3  cos  60°  =2.4  X6.3  Xsin  30°.  The 
setting  requires  the  A,  B,  and  S  scales,  as  follows: 

A2A  <    ao        n  ■       ,c,  7.56  J. 

- — -  =runner  to  6.3  on  B  .  .  .  rin  of  >S  to  runner  =  . 

lin  B  30  o 

5.7 
Ex.    185.     Determine   <{>   from   the   equation  tan  $=777:.   Set 

0.8 

6.8  on  C  over  5.7  on  D;  move  runner  to  right  index  of  C;   under 

runner  read  40°  on  T. 

The  following  Table  VII  shows  the  method  of  determining 
the  worth  of  a  continued  product  or  the  worth  of  a  com- 
pound ratio. 


118 


PEACTICAL  MATHEMATICS 


a 

[ 
I 

< 

00 

a 

o 

00 
CO 

oq 

CO 
CO 

© 

00 
00 

U    72 

1—1 

CO 

"* 

o 

1— 1 

1 

- 

g 

3 

a> 

^2 

CO 

8 
| 

H 

h 
3 

£ 

s 

o 

+3 

rH 

1—1 

i—i 

1—1 

> 

1— 1 

"3 

o 

s 

S 

Q 

/■ — V 

i— i 

"o 

u 

02   (4 

E  o 
©hi 

Q 

o 

1 

o 

d 

m 

a 

3 

u 

o 

03 

a 

1 

o 

a 

02 

Q 

1—1 

<N 

cq 

o 

"S 

K 

02 

S 
9 
2 

<M 

<* 

*o 

- 

O 

<N 

a 

8 

'c 
E 
a 

► 

i 

CO 

X 

IC 

t^ 

X 
<m 

i—i 

X 

00 

CO 
CO 

X 

00 

X 

d 
X 

00 
CO 
•1- 

id 

T— 1 
"I' 

CO 
CO 

00 

X 

i—i 

OS 

CO 

X 

CO 

FUNDAMENTAL  OPERATIONS 


119 


Ex.  186.     Use  the  slide  rule  to  perform  the  following  examples 
as  a  check  on  the  preceding  work : 


(a)  Ex.  16, 

(b)  Ex.  31, 

(c)  Ex.33, 

(d)  Ex.  80-82, 

(e)  Ex.  of  areas  85, 
(/)   Ex.  92-93, 

iff)  Ex.  97-101, 
(h)  Ex.  102, 
(i)   Ex.  103-105, 
(j)   Ex.  110-113, 
(k)  Ex.  114, 


(I)   Ex.  115-120, 
(w)  Ex.  126, 
(?i)  Ex.  152, 
(o)  Ex.  154-156, 
(p)  Ex.  157, 
(q)  Ex.  158-159, 
(r)   Ex.  160-161, 
(s)  Ex.  162-163, 
(0    Ex.  164-165, 
(u)  Ex.  166, 
(v)  Ex.  167-169. 


By  means  of  the  slide  rule  show  that: 


(w) 


1 

2 

sin  0F^sin 

e 

2 

1 
3 

tan  6?^tan 

6 

3~ 

2 

cos  0^cos 

26 ; 

test  these  for  ten  values  of  0. 


1  N 

(x)  —log  iV?^  log  — .     Test  this  for  ten  values  of  N. 

*    —  ^ 

(2/)  log  (ffl+&)  5^  log  a+log  6;  test  this  for  ten  values  of  a  and  b. 
(2)  sin  (8+<{))  5^  sin  0  +  sin  cj>;  test  this  for  ten  values  of  0  and  <]>. 

Observation.  A  multiplier  of  a  function  must  precede 
f  he  function  name  and  not  the  quantity.  A  divisor  of  a  function 
must  be  written  under  the  function  name  and  not  under  the 
quantity. 

A  function  name  extends  only  to  the  term  or  parenthesis 
immediately  following  it.  The  function  name  cannot  be  dis- 
tributed, i.e.,  written  separately  before  the  terms  included 
in  a  parenthesis. 

Ex.  187.  Prepare  a  table  giving  the  following  forms,  their 
numeric  values  and  their  logarithms. 


120  PRACTICAL  MATHEMATICS 

(a)  The  nine  multiples  of  x. 

(b)  The  first  five  powers  x. 

(c)  x  divided  by  the  following  respective  divisors:   2,  3,  4,  5,  6, 

7,  8,  9,  12,  16,  32,  64,  108,  180 

(d)  The  first  five  roots  of  %. 

(e)  xVi",   %V^t  4x2,   x2^-4,   xV2^  x^-V2^   2V%,    V2x",    \Q" 

V2-x,  V3-x,  V4-x,  V90-X,  V4x-=-3* 

(/)   ^2x,  ^xT2,  ^2Ti,  V^T4,  VzVrt  ^6Tx,  ^x2,  xvV. 
fe)  0,  9*,  V9,  1-0,  V2sr,  xVgr,  W2g,  %+g,  %  +  V2g,  x*+g. 
(h)   e,  s2,  ««,  e4,  1  -S-e,  1  -e2,  1  -e3,  1  -e4,  V^  ^7. 

The  following  settings  for  the  C  and  D  scales  are   of 
frequent  use : 

set  226  _      diameter  circle 
to  710     circumference  circle, 
99  diameter  circle 


70    side  of  inscribed  square, 

26  _      inches 

66     centimeters, 

79        diameter  circle 


70    side  of  equal  square, 

39  _  circumference  circle 

11     side  of  equal  square, 
31  sq.  inches 

200     sq.  centimeters, 

To  obtain  compound  interest,  set  the  left  index  of  C 
opposite  (1+rate)  on  LL  1,  then  the  ratio  of  amount  to 
principal  is  read  on  LL  2,  opposite  the  number  of  years 
on  C.     Thus  a  principle  will  double  in  14  years  at  5  per  cent. 

LLIL05  =  LL_22 

C  lin  14  C 

Ex.  188.  Show  that  a  principal  at  6  per  cent  compound 
interest  will  double  in  11  years  10  mos.  21  days. 


FUNDAMENTAL  OPERATIONS  121 

TABLE  VIII.     TRIGONOMETRIC    FUNCTIONS 


Angle. 

Chord. 

X.F. 

Sine. 

<f>  F. 
Tangent. 

Cotan. 

R.  F. 

Cosine 

1 

De- 

i 
Radians 

grees 

0° 

0 

0 

0 

0 

00 

1 

1.414 

1 . 5708 

90° 

1 

.0175 

.017 

.0175 

.0175 

57.2900 

.9998 

1.402 

1.5533 

89 

2 

.0349 

.035 

.0349 

.0349 

28.6363 

.9994 

1.389 

1 . 5359 

88 

3 

.0524 

.052 

.0523 

.0524 

19.0811 

.9986 

1.377 

1.5184 

87 

4 

.0698 

.070 

.0698 

.0699 

14.3007 

.9976 

1.364 

1.5010 

86 

5 

.0873 

.087 

.0872 

.0875 

11.4301 

.9962 

1.351 

1.4835 

85 

6 

.1047 

.105 

.1045 

.1051 

9.5144 

.9945 

1.338 

1.4661 

84 

7 

.1222 

.122 

.1219 

.1228 

8.1443 

.9925 

1.325 

1.4486 

83 

8 

.1396 

.140 

.1392 

.1405 

7.1154 

.9903 

1.312 

1.4312 

82 

9 

.1571 

.157 

.1564 

.1584 

6.3138 

.9877 

1.299 

1.4137 

81 

10 

.1745 

.174 

.1736 

.  1763 

5.6713 

.9848 

1.286 

1.3963 

80 

11 

.1920 

.192 

.1908 

.1944 

5.1446 

.9816 

1.272 

1.3788 

79 

12 

.2094 

.209 

.2079 

.2126 

4.7046 

.9781 

1.259 

1.3614 

78 

13 

.2269 

.226 

.2250 

.2309 

4.3315 

.9744 

1.245 

1.3439 

77 

14 

.2443 

.244 

.2419 

.2493 

4.0108 

.9703 

1.231 

1.3265 

76 

15 

.2618 

.261 

.2588 

.2679 

3.7321 

.9659 

1.218 

1 . 3090 

75 

16 

.2793 

.278 

.2756 

.2867 

3.4874 

.9613 

1.204 

1.2915 

74 

17 

.2967 

.296 

.2924 

.3057 

3 . 2709 

.9563 

1.190 

1.2741 

73 

18 

.3142 

.313 

.3090 

.3249 

3.0777 

.9511 

1.176 

1.2566 

72 

19 

.3316 

.330 

.3256 

.3443 

2.9042 

.9455 

1.161 

1.2392 

71 

20 

.3491 

.347 

.3420 

.3640 

2.7475 

.9397 

1.147 

1.2217 

70 

21 

.3665 

.364 

.3584 

.3839 

2.6051 

.9336 

1.133 

1 . 2043 

69 

24 

.3840 

.382 

.3746 

.4040 

2.4751 

.9272 

1.118 

1 . 1868 

68 

23 

.4014 

.399 

.3907 

.4245 

2.3559 

.9205 

1.104 

1.1694 

67 

24 

.4189 

.416 

.4067 

.4452 

2.2460 

.9135 

1.089 

1.1519 

66 

25 

.4363 

.433 

.4226 

.4663 

2.1445 

.9063 

1.075 

1.1345 

65 

26 

.4538 

.450 

.4384 

.4877 

2.0503 

.8988 

1.060 

1.1170 

64 

27 

.4712 

.467 

.4540 

.5095 

1.9626 

.8910 

1 .  045 

1.0996 

63 

28 

.4887 

.484 

.4695 

.5317 

1.8807 

.8829 

1.030  1.0821 

62 

29 

.5061 

.501 

.4848 

.5543 

1.8040 

.8746 

1.015 

1.0647 

61 

30 

.5236 

.518 

.5000 

.5774 

1.7321 

.8660 

1.000 

1.0472 

60 

31 

.5411 

.534 

.5150 

.6009 

1 . 6643 

.8572 

.985 

1.0297 

59 

32 

.5585 

.551 

.5299 

.6249 

1 . 6003 

.8480 

.970 

1.0123 

58 

33 

.5760 

.568 

.5446 

.6494 

1 . 5399 

.8387 

.954 

.9948 

57 

34 

.5934 

.585 

.5592 

.6745 

1.4826 

.8290 

.939 

.9774 

56 

35 

.6109 

.601 

.5736 

.7002 

1.4281 

.8192 

.923 

.9599 

55 

36 

.6283 

.618 

.  5878 

.7265 

1 . 3764 

.8090 

.908 

.9425 

54 

37 

.6458 

.635 

.6018 

.7536 

1 . 3270 

.7986 

.892 

.9250 

53 

38 

.6632 

.651 

.6157 

.7813 

1.2799 

.7880 

.877 

.9076 

52 

39 

.6807 

.668 

.6293 

.8098 

1 . 2349 

.7771 

.861 

.8901 

51 

40 

.6981 

.684 

.6428 

.8391 

1.1918 

.7660 

.845 

.8727 

50 

41 

.7156 

.700 

.6561 

.8693 

1 . 1504 

.7547 

.829 

.8552 

49 

42 

.7330 

.717 

.6691 

.9004 

1.1106 

.7431 

.813 

.8378 

48 

43 

.7505 

.733 

.6820 

.9325 

1.0724 

.7314 

.797 

.8203 

47 

44 

.7679 

.749 

.6947 

9657 

1 . 0355 

.7193 

.781 

.8029 

46 

45° 

.7854 

.765 

.7071 

1.0000 

1 . 0000 

.7071 

.765 

.7854 

45° 

Radians 

De- 
grees. 

Cosine 

Co-tan. 

Tangent. 

Sine. 

Chord. 

R.  F. 

<£  F. 

X.  F. 

Ang 

le. 

122 


PEACTICAL  MATHEMATICS 


TABLE  IX.     LOGARITHMS 


N 

The  Third  Figure  of  Your  Number. 

Proportional  Parts. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

12  3  4 

o 

6 

7 

8  9 

10 

0000 

0043 

0086 

0128 

0170 

4  9  13  17 

21 

26 

30  34  38 

0212 

0253 

0294  0334 

0374 

4  8  12  16 

20 

24 

28 

32  37 

11 

0414 

0453 

0492 

0531 

0569 

1 

4  8  12  15 

19 

23 

27 

31  35 

0607 

0645 

0682 

0719 

0755 

4  7  11  15 

19 

22 

26 

30  33 

12 

0792 

0828 

0864 

0899 

0934 

0969 

3  7  11  14 

18 

21 

25 

28  32 

1004 

1038 

1072 

1106 

3  7  10  14 

17 

20 

24 

27  31 

13 

1139 

1173 

1206 

1239 

1271 

3  7  10  13 

16 

20 

23 

26  30 

1303 

1335 

1367 

1399 

1430 

3  7  10  12 

16 

19 

22 

25  29 

14 

1461 

1492 

1523 

1553 

3  6 

9  12 

15 

18 

21 

24  28 

1584 

1614 

1644 

1673 

1703 

1732 

3  6 

9  12 

15 

17 

20 

23  26 

15 

1761 

1790 

1818 

1847 

1875 

1903 

36 

9  11 

14 

17 

20 

23  26 

1931 

1959 

1987 

2014 

35 

8  11 

14 

16 

19 

22  25 

16 

2041 

2068 

2095 

2122 

2148 

35 

8  11 

14 

16 

19 

22  24 

2175 

2201 

2227 

2253 

2279 

3  5 

8  10 

13 

15 

18 

21  23 

17 

2304 

2330 

2355 

2380 

2405 

2430 

35 

8  10 

13 

15 

18 

20  23 

2455 

2480 

2504 

2529 

|2  5 

7  10 

12 

15 

17 

19  22 

18 

2553 

2577 

2601 

2625 

2648 

25 

7  9 

12 

14 

16 

19  21 

2672 

2695 

2718 

2742 

2765 

25 

7  9 

11 

14 

16 

18  21 

19 

2788 

2810 

2833 

2856 

2878 

24 

7  9 

11 

13 

16 

18  20 

2900 

2923 

2945 

2967 

2989 

24 

6  8 

11 

13 

15 

17  19 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

24 

6  8 

10 

12 

14 

16  18 

22 

3424 

3444 

3464 

348313502 

3522 

3541 

3560 

3579 

3598 

24 

6  8 

10 

12 

14 

15  17 

23 

3617 

3636 

3655 

3674  3692 

3711 

3729 

3747 

376J3 
3945 

3784 

24 

6  7 

9 

11 

13 

15  17 

24 

3802 

3820 

3838 

3856  3874 

3892 

3909 

3927 

3962 

24 

5  7 

9 

11 

12 

14  16 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

23 

5  7 

9 

10 

12 

14  15 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

23 

5  7 

8 

10 

11 

13  15 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

23 

5  6 

8 

9 

11 

13  14 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

23 

5  6 

8 

9 

11 

12  14 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

1  3 

4  6 

7 

9 

10 

12  13 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

1  3 

4  6 

7 

9 

10 

11  13 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

1  3 

4  6 

7 

8 

10 

11  12 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

1  3 

4  5 

7 

8 

9 

11  12 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

1  3 

4  5 

6 

8 

9 

10  12 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

1  3 

4  5 

6 

8 

9 

10  11 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

5611 

5623 

5635 

5647 

5658 

5670 

1  2 

4  5 

6 

7 

8 

10  11 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5775 

5786 

1  2 

3  5 

6 

7 

8 

9  10 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

5877 

5888 

5899 

39 

5911 

5922 

5933 

5944 

5955 

5966 

5977 

5988 

5999 

6010 

1  2 

3  4 

5 

7 

S 

9  10 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

1  2 

3  4 

5 

6 

8 

9  10 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

1  2 

3  4 

5 

6  7 

8  9 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

6618 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

1  2 

3  4 

5 

6 

7 

7  8 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

1  2 

3  4 

5 

5 

6 

7  8 

48 

6812 

6821 

6830 

6839 

5848 

6857 

6866 

6875 

6884 

6893 

1  2 

3  4 

4 

5 

6 

7  8 

49 

6902 

6911 

6920 

6928 

6937 

594  6 

6955 

6964 

6972 

6981 

50 

6990 

6998 

7007 

7016 

702417033 

7042 

7050  7059 

7067 

12 

3  3 

4 

5 

6 

7  8 

FUNDAMENTAL  OPERATIONS 


123 


TABLE  IX.    LOGARITHMS 


7853 
7924 
7993 

8062 
8129 


Sli 


66 

67  18261  8267 

68  8325  8331 


70 


8287  8293  8299 


18202!  8209  8215 

8274  8280 

8338  8344;8351|8357  8363 
8388  8395  8401  8407  8414  8420  8426 
8451  845718463  8470|8476  8482  8488 


90 


The  Third  Figure  of  Your  Number. 


076  7084  7093  7101,7110 
7160  7168  7177 
51  7259 
7332 1 7340 


7243172 

324 
7404 


7482 
7559 
7634 
7709 

7782 


78607868 
793117938 
8000:8007 
8069|8075 
81368142 


513  8519 
573  8579 


8633 
8692 
8751 


76  I880S 

77 

78 

79  18976  8982 

80  9031  9036 


8865  8871 
8921  8927 


934.: 
9395 


9542 


£11 


7412:7419 


7185,7193  7202  7210 
7267J7275  7284  7292 
7348  7356  7364  7372 
7427i7435  7443  7451 


7490  7497  7505,75 
7566  7574 |7582j7589 
7642  7649  7657i7664 
7716  7723  7731  7738 

7789  7796  78037810  7818 


13  7520  7528 
7597  7604 
7672  7679 
7745 


7875 1 7882 
7945  7952 
8014  8021 
8082  8089  8096 
8149,8156  8162 


8222  8228 


8639 

8698 
8756 


8525;8531| 
8585  8591] 
8645  8651  j 
87048710 

8762  8768 


E=E 


8814  8820  8825 


887618882 
8932  8938 
8987  8993 
9042  9047 


8831 
8887 
8943 
8998 
9053 


81  9085  9090|9096  9101 

82  9138  9143  9149  9154 

83  9191  9196:9201  9206 

84  9243  9248  9253  9258  926319269 

85  |9294|9299  930493099315|9320|9325 


s^sogssstoo'oses  9370  9375 

9400  9405  9410  9415 
9445  9450  9455  9460  9465 
89  |9494  9499  9504  9509  9513 
9547  9552i9557i95 


9590  9595  9600  9605'9609|9614|9619 
9638  9643  9647:9652  9657  9661  9666 
9685  9689  9694 !9699  9703  9708  9713 
9731  9736  9741  9745  9750  9754  9759 
9777  9782  97869791  9795  9800  9805 


9827  9832  9836984119845 

9872,9877  9881 19886  9890 

|9912|9917  9921  9926  9930  9934 

19965  9969  9974|9978 


9823 
9868 


9956  9961 


7118 


7889 
7959 
8028 


8537  8543  8549 
8597  8603  8609 
8657|8663  8669 

8727 


8716  8722 
8774  8779 


8837 
8893 
8949 
9004 
9058 


9106  9112 
9159  9165 
9212  9217 


9420  9425 
9469  9474 
9518  9523 
62  9566  9571 


7126 


7752 
7825 


7896 
7966 
8035 
8102 
8169 


8235 


8785 


8842 
8899 
8954 
9009 
9063 


9117 
9170 
9222 
9274 


9850 
9894 
9939 
9983 


7135 

7218 
7300 
7380 
7459 


7536 
7612 
7686 
7760 
7832 


7903 
7973 
8041 
8109 
8176 


8241 
8306 
8370 
8432 
8494 


8555 
8615 
8675 
8733 
8791 


8848 
8904 
8960 
9015 
9069 


9122 
9175 
9227 
9279 
9330 


9380 
9430 
9479 
9528 
9576 


9624 
9671 
9717 
9763 
9809 


9854 


9943 
9987 


S 


7143 
7226 
7308 
7388 
7466 


7543 
7619 
7694 
7767 
7839 


7910 
7980 
8048 
8116 
8182 


8248 
8312 
8376 
8439 
8500 


8561 
8621 
8681 
8739 
8797 


8854 
8910 
8965 
9020 
9074 


9128 
9180 
9232 
9284 
9335 


9385 
9435 
9484 
9533 
9581 


9628 
9675 
9722 
9768 
9814 


9 


7152 
7235 
7316 
7396 
7474 


7551 
7627 
7701 

7774 
7846 


7917 
7987 
8055 
8122 
8189 


8254 
8319 
8382 
8445 
8506 


8567 
8627 


8745 
8802 


8859 
8915 
8971 
9025 
9079 


9133 
9186 
9238 
9289 
9340 


9390 
9440 
9489 
9538 
9586 


9633 
9680 
9727 
9773 
9818 


9859  9863 
9903  9908 
9948  9952 
9991  9996 


Proportional  Parts. 


12    3    4 


12  3  3 

12  2  3 

12  2  3 

12  2  3 


112    3 
112    3 


112    3 


112    3 
112    2 


112    2 


112    2 


0  112 


0112 


6     7     8     9 


5  6 

5  6 

5  6 

5  5 


4     5     6     7 
4     5     6     6 


4     5     5     6 


4     4     5 
4     4     5 


4     4     5     5 


3     4     4     5 


3     3     4     4 


3     3     3     4 


CHAPTER  II 
THE  TRANSFORMATION  OF  FORMULAS 

The  Formula.  A  formula  is  a  condensed  form  of  the 
verbal  statement  of  a  law  of  science. 

In  1827,  G.  S.  Ohm  annunciated  his  celebrated  law: 
the  intensity  or  strength  of  an  electric  current,  i.e.,  the 
quantity  of  electricity  passing  a  section  of  the  conductor 
in  a  unit  of  time,  is  directly  proportional  to  the  whole 
electromotive  force  in  operation,  and  inversely  proportional 
to  the  sum  of  all  the  resistances  in  the  circuit. 

Ohm's  Law  may  be  abbreviated  as  follows : 

-,-,.-  .   ;  N     Electromotive  force  (volts) 

Intensity  of  current  (amperes)  = ^    .  . ,  .     \ -. 

Resistance  (ohms) 

In  other  words,  the  intensity  of  the  current  is  equal  to 
the  quotient  obtained  by  dividing  the  electromotive  force 
by  the  resistance. 

The    Notation.     A   group   of  abbreviations  with   their 
meanings  is  called  the  notation. 
Notation  : 

Let  E  =  the  whole  electromotive  force  in  operation, 
/  =  the  intensity  of  the  current, 
R  =  the  sum  of  the  resistances  in  the  circuit. 

By  the  use  of  the  above  notation  the  formula  becomes : 

I-* 

R 

The  Data.  The  values  which  have  been  assigned  to 
the  notation  of  a  formula  are  called  the  data.     The  data 

124 


THE  TRANSFORMATION  OF  FORMULAS  125 

should   include  the  numeric  values  of  all  letters  but  one. 
That  letter  may  be  computed  for  its  numeric  value. 

Data: 

#  =  0.05  volts  =  50  millivolts, 

/  =  200  amperes, 

R  =  ?  (the  value  of  R  is  to  be  determined.) 

The  Transformation.  The  transformation  of  a  formula 
is  the  result  of  performing  one  or  more  operations  upon 
a  formula  in  order  to  obtain  the  literal  solution  of  any 
of  its  letters. 

Transformation  for  R       mathematic  authority 

E 

(1)  /  =  -= quotation  of  Ohm's  Law, 

ti 

(2)  RI  =  E multiplying  (1)  by  R  (mul  ax), 

(3)  R  =  j dividing  (2)  by  J  (div  ax). 

The  Solution.  The  data  should  include  the  numeric 
values  of  all  letters  but  one.  That  one  letter  is  said  to  be 
an  unknown  and  the  determination  of  its  numeric  value 
is  called  the  solution  of  the  formula. 

The  data  may  be  substituted  in  (1),  giving  (4),  which 
leads  to  (5),  or  by  substituting  the  data  in  (3),  which 
gives  (5)  directly. 

(4)  200=^, 

(5)  ^=255, 

simplifying  (5)  we  obtain  (6) 

(6)  R  =  .00025  ohm  =  250  microhms. 

The  Interpretation.  The  notation  enables  us  to  make 
formula  work  a  reversible    process.     A  formula  is   inter- 


126  PEACTICAL  MATHEMATICS 

preted  as  a  verbal  law  by  reading  into  the  abbreviations 
their  significance  as  expressed  in  the  notation. 

The  same  current  passes  through  every  cross-section  of 
a  circuit,  so  that  if  we  consider  any  length  of  the  circuit 
we  can  use  Ohm's  Law  providing  we  know  the  difference 
of  voltage  between  the  ends  of  a  conductor  and  the  corre- 
sponding resistance.  Therefore,  for  a  segment  or  any 
definite  length  of  a  circuit  or  conductor  we  use  the  fol- 
lowing notation : 

E  =  the  difference  of  potential  (P.D.),  i.e.,  the  voltage  dif- 
ference or  so  called  voltage  drop  between  two 
points  in  an  electric  circuit  or  conductor, 

ii^the  corresponding  resistance  between  the  two  points, 
J  =  the  intensity  of  the  current  flowing  through  the  circuit 
or  conductor. 

Interpretation  of  formula  (2) : 

(2)  E  =  IR  consists  of  three  distinct  elements — 

[]  =  ()!  }• 

[The  voltage  drop  or  potential  difference  between  any 
two  points  of  a  conductor]  equals  the  product  of  {the  resist- 
ance of  the  conductor!  times  (the  current  flowing  through  it). 

Interpretation  of  formula  (3) : 

rp  r     1 

(3)  R  —  j  consists  of  three  distinct  elements.     J  |.=  — -• 

{The  resistance  of  a  conductor!  is  equal  to  the  ratio  of 
[the  voltage  drop  across  the  conductor]  to  (the  current 
passing  through  it) . 

The  Analysis  of  a  Formula.  The  outline  of  the  above 
example  is  called  an  analysis  and  represents  the  mental 
and  written  processes  which  should  be  applied  to  every 
problem.     The  following  order  of  procedure  suggests  the 


THE  TRANSFORMATION  OF  FORMULAS         127 

method  of  attacking  a  problem:  (a)  at  the  beginning 
enter  the  heading  which  is  the  title  or  subject  of  the  formula; 
(6)  under  the  heading  write  the  formula  or  formulate  the 
given  law  or  problem;  (c)  on  the  right-hand  side  of  the 
page  state  the  notation,  and  the  (d)  data;  under  the  formula 
enter  a  series  of  numbered  statements  representing  the  (e) 
transformations  of  the  formula;  (/)  the  numeric  sub- 
stitution of  the  data;  (g)  the  simplification  of  the  result; 
(h)  the  interpretation  of  the  formula  in  its  various  trans- 
formations may  follow. 

Observation.  The  formulation  of  a  law  or  problem  is  the 
focusing  or  concentration  of  facts  by  the  use  of  abbreviations 
and  symbols  which  also  express  the  relations  between  things 
and  the  forces  which  act  upon  them,  and  thereby  enable  the 
mind  to  appreciate  a  law  in  its  entirety. 

Induced  E.M.F.  in  a  moving  coil.  The  E.M.F.  which 
is  induced  in  a  coil  moving  through  a  magnetic  field  is 
proportional  to  the  rate  of  cutting  the  magnetic  flux,  i.e., 
the  number  of  lines  of  force  cut  per  second. 

An  equivalent  statement  would  be:  the  electromotive 
force,  in  volts,  generated  in  a  moving  coil  is  equal  to  the 
product  of  the  total  flux  cut  times  the  number  of  revo- 
lutions per  second  divided  by  the  proportionality  factor  108. 

Voltage  generated  in  a  coil : 

Notation: 
(  )  F  =  —  #  =  E.M.F.  generated  in  the  coil, 

108'  <£  =  total  flux  in  the  magnetic   circuit 

(b)  $=AH,  which  is  cut  by  the  coil, 

(c)  A  =  Id,  N  =  the  number  of  revolutions  per  second 

of  the  coil, 
A  =  area  of  the  coil  in  sq.cm., 
H  =  the  strength  of  the  field  in  gausses, 
per  sq.cm., 
Z  =  the  length  of  the  coil  in  cm. 
d  =  the  breadth  of  the  coil  in  cm. 


128 


PRACTICAL  MATHEMATICS 


Data: 
E  =  ? 

#  =  10000  gausses, 
Z  =  60cm., 
iV=180r.p.m.  =  3  r.p.s., 
d  =  30  cm. 

(a),  (b),  and  (c)  are  simultaneous  equations,  i.e.,  equations 
in  which  the  same  letters  have  the  same  significance  and 
the  same  numeric  values. 

The  solution  for  E  may  be  performed  in  one  of  two  ways : 
substitute  the  value  of  A  from  (c)  in  (b)  in  order  to  obtain 
3>,  then  substitute  the  latter  in  (a),  which  gives  the  trans- 
formed value  of  E  in  terms  H,  N,  I,  d,  the  numeric  values 
of  which  are  then  substituted ;  another  method  of  proceed- 
ing is  to  substitute  the  data  and  calculated  values  in  each 
of  Eqs.  (c),  (b),  and  (a)  in  succession.  In  both  cases  we 
arrive  at  the  same  numeric  value  for  E. 


First  Method. 
(6)  $  =  AH, 

(c)  A  =  ld, 

(1)  *=ldH, 

(a)  E=^s, 


(2)  E 


(3)  E  = 


IdHN 
108  ' 

60X30X10000X3 


108 


(4)  #=.54  volt. 


Second  Method. 
(c)  A  =  ld, 

(1)  A  =  60X30=  1800, 
(6)    $  =  AH, 

(2)  $=1800X10000=18X106, 


(3)  E  = 


18X106X3 


108 
(4)  #=.54  volt. 


Interpretation  of  formula  (6) : 

The  total  flux  in  a  magnetic  circuit  equals  the  product 
of  the  area  times  the  strength  of  the  field. 


THE  TRANSFORMATION  OF  FORMULAS  129 

Interpretation  of  formula  (c)  : 

The  area  of  the  coil  equals  the  product  of  its  length 
times  its  breadth. 

By  transforming  (a)  we  obtain  (5)  and  (6) : 

(5)  -^, 

Interpretation  of  (5) : 

The  total  flux  required  to  generate  a  given  E.M.F. 
equals  108  times  the  E.M.F.  divided  by  the  number  of 
revolutions  per  second. 

Interpretation  of  (6) : 

The  number  of  revolutions  per  second  required  to  generate 
a  given  E.M.F.  equals  108  times  the  E.M.F.  divided  by  the 
total  flux. 

Transformation  of  Formulas.  The  transformation  of  a 
formula  implies  the  necessary  steps  through  which  it  is 
modified  to  give  the  solution  for  its  letters.  Every  step 
must  be  assured  by  definite  mathematic  authorities. 

A  test  for  the  accuracy  of  transformation  work  is 
reversion  to  the  initial  formula. 

The  solution  of  a  formula  requires  a  letter  isolated  in 
one  member  of  the  equation,  so  as  to  have  a  coefficient 
and  exponent  of  unity. 

Transform  the  following  formulas  and  express  the  solution 
for  each  letter.  Show  authorities  and  then  present  the  work  in 
the  form  which  is  in  accordance  with  the  standards  set  for  this 
department. 

Ex.1.    Q=IL  Solve  for  I,t. 

Ex.  2.     I=~  Solve  for  W,  K2,  t. 

K21 

Ex.  3.     I  -— .  Solve  for  V,  Kl}  t. 

Kit 


130  PRACTICAL  MATHEMATICS 

Vh  273 
EX-i-    I=  0.1733  X76(273  +  Df    So1™  ioT  V>  »' L 
Divide  (3)  by  (4).  Solve  for  K  in  terms  of  h. 

Ex.5.    C«|(F -32J  Solve  fori?7. 

Ex.  6.     72  -  — .  Solve  for  K,  L,  d. 

Ex.  7.     Ohm  =  1000000  microhms.   Express  microhms  in  ohms. 

_     _      JL         megohm  _  _        .      . 

Ex.  8.    Ohm  =  .  Express  megohms  in  ohms. 

1 000000 

Ex.9,    d2    =CM.  Solve  ford. 

Ex.  10.    sq.  mils  =  CM  0.7854.        Solve  for  CM. 

There  are  a  few  cases  where  two  letters  are  used  to  abbreviate 
a  quantity,  as  illustrated  in  (9)  and  (10).  In  such  cases,  these 
compounded  letters  are  to  be  treated  as  inseparable. 


CHAPTER  III 

INTERPRETATION  OF  FORMULAS 

For  the  formula  work  of  this  chapter  use  the  notation 
given  below.  Interpret  the  formulas  and  complete  those 
statements  of  laws  which  are  stated  in  part. 

Z=time  in  seconds  which  elapses, 
Q  =  quantity  of  electricity  passing  a  point  in  a  cir- 
cuit in  coulombs, 
J  =  rate  of  flow,  or  intensity  of  current  in  amperes, 
W  =  weight  in  grams  gained  or  deposited  in  a  volta- 
meter, 
2^2  =  weight  in  grams  deposited  by  one  amp.  per  sec. 
V  =  volume  in  cu.cm.  evolved  in  a  gas  voltameter, 
Ki  =  volume  in  cu.cm.   evolved    in  a  gas  voltameter 
by  one  amp.  per  sec, 
h  =  pressure  of  the  atmosphere  in  cm., 
R  =  resistance  of  a  conductor  in  ohms, 
L  =  length  of  a  conductor  in  feet, 
d  =  diameter  of  a  conductore  in  mils, 
K  =  resistance  of  a  mil-foot  of  wire, 
CM  =  circular  mils, 

T= temperature  of  a  room  in  degrees  Centigrade. 

In  many  cases  a  quantity -is  abbreviated  by  its  initial 
letter  and  in  other  cases  where  a  conflict  might  arise  by  an 
arbitrarily  chosen  letter.  Wherever  possible  the  notation 
of  convention  is  adopted  in  this  text.  /  in  this  instance 
is  the  initial  letter  in  the  expression  "  intensity  of  current," 
because  C  is  used  as  the  abbreviation  for  capacity. 

131 


132  PRACTICAL  MATHEMATICS 

Ex.  1.  t=j,     from  (1) 

using  the  notation 

quantity  (coulombs) 

time  (seconds)  = ~ — ; -. 

current  strength  (amperes) 

/.  The  law  to  find  the  time  (in  seconds)  required  for  a  given 
quantity  of  electricity  (in  coulombs)  to  pass  a  point  in  a  circuit, 
divide  the  quantity  of  electricity  (in  coulombs)  by  the  rate  of 
flow  (in  .  .  .  ). 

Ex.  2.    Interpret  Q  =tl  as  follows: 

Law  for  the  quantity  of  electricity  (in  coulombs)  to  pass  a 

point  in  a  circuit :  .  .  .  ? 

W 
Ex.  3.     Transform  I  =—  for  t  and  interpret  as  follows: 
K2t 

Law  for  the  time  required  to  deposit  electrically  a  given  weight 

of  metal  with  a  given  current  strength? 

V 
Ex.  4.    Interpret  I  =—  as  follows: 
Kit 

Law  to  calculate  strength  of  an  unknown  current  when  a 
volume  of  gas  is  evolved  in  a  given  time :  .  .  .  ? 

Ex.  5.  Transform  the  formula  in  Ex.  3  for  W  and  interpret 
as  follows: 

Law  for  the  weight  of  any  metal  that  will  be  deposited  in 
a  voltameter  by  a  given  current  in  a  given  time :  .  .  .  ? 

Ex.  6.  Transform  the  formula  in  Ex.  4  for  V  and  interpret 
as  follows: 

Law  for  the  volume  of  a  gas  generated  by  a  known  current 
in  a  given  time:  .  .  .  ? 


CHAPTER  IV 

THE   FORMULATION   OF   PROBLEMS 

The  Ampere-hour.  The  coulomb  is  a  very  small  quan- 
tity of  electricity.  For  commercial  purposes  a  larger  unit, 
the  ampere-hour,  is  used. 

One  ampere-hour  is  the  quantity  of  electricity  that 
would  pass  any  point  in  a  circuit  in  one  hour  when  the 
strength  of  current  is  one  amp.  An  equivalent  would  be 
2  amps,  flowing  for  J  hour. 

Ex.  1.  What  is  the  equivalent  of  one  amp.-hour  (a)  for  a 
current  flowing  for  4  hours;  (b)  \  ampere  requires  what  equivalent 
time;  (c)  show  that  3600  coulombs  is  the  equivalent -of  an  amp.- 
hour. 

_      _  ampere-hours      _  J  „.     ,         . 

Ex.  2.    Amperes  = ■ .    Interpret    this  formula  as 

hours 

follows: 

Law  for  ampere-hours  consumed  in  an  electric  circuit :  multiply 
the  average  strength  of  current  (in  ?)  by  the  ? 

Ex.  3.  A  current  of  6.5  amp.  was  maintained  by  a  cell  for  5 
hours.  What  quantity  of  electricity  has  been  used?  Express  in 
coulombs  and  also  in  ampere-hours  (amp.-hrs.). 

Ex.  4.  Suppose  the  cell  in  Ex.  3  has  a  capacity  of  80  amp.- 
hrs.,  how  long  could  the  above  current  be  maintained? 

Use  the  formula  in  Ex.  2. 

Ex.  5.  How  long  a  time  will  be  required  to  deposit  5  grams 
of  silver  on  a  copper-plated  teaspoon  with  a  current  of  2  amps.? 

K2  for  silver  =  0.001118  gram. 

Ex.  6.  The  negative  plate  of  a  copper  voltameter  has  increased 
its  weight  1.9  grams  in  35  seconds.  What  was  the  average  rate 
of  current  strength?     K2  for  copper  -  0.0003286. 

Ex.  7.  In  an  electroplating  bath  how  many  grams  of  zinc 
will  be  deposited  by  a  current  of  5  amps,  in  60  min?  K2  for  zinc 
0.0003386. 

133 


134  PRACTICAL  MATHEMATICS 

Ex.  8.  The  following  data  are  recorded  in  a  gas  voltameter 
test: 

Level  of  burette  before  test  2.7  cu.cm. 

Level  of  burette  after  test,  32.2  cu.cm. 

Vol.  of  gas  evolved  =32.2  -2.7  =? 

Time  of  closing  switch  8  hr.  40  min.  15  sec. 

Time  of  opening  switch  8  hr.  45  min.  15  sec. 

Length  of  run  =?  min.  =?  sec? 

Total  gas  generated  per  sec.  =? 

One  amp.  in  one  sec.  (one  coulomb)  generates  0.1733  cu.cm. 
gas  per  sec.    Substitute  in  the  formula: 

""•  W 

Ex.  9.  For  more  accurate  determinations  Formula  (4)  from 
Chapter  II  is  used.  In  an  experiment  the  volume  of  gas  generated 
in  a  gas  voltameter  was  found  to  be  25  cu.cm.  in  65  sec,  its  tem- 
perature (taken  as  temperature  of  the  room)  was  20°  centigrade, 
and  the  atmospheric  pressure  was  equal  to  76  cm.  What  was 
the  current  strength? 

Ex.  10.  A  piece  of  sheet  iron  6  ins.  square  is  to  be  plated  on 
both  sides  in  a  copper  acid  bath.  What  current  strength  is  re- 
quired? 

Current  density  for  Cu  cyanides  5-10  amp.  per  sq.ft. 
Allow  7.5  amps,  per  sq.ft. 
Area  of  plate  both  sides  =  ?  sq.ft. 
7X?=?  amps. 
Ex.  11.    Give  the  source  of  information  and  show  range  of 
variation  of  values  of  the  current  strength  used  in  practice  to 
operate  the  following: 

(a)  Current  required  to  operate  a  110-volt  16  cp.  inc.  lamp. 
(6)  Current  required  to  operate  an  enclosed  arc  lamp  1 10-volt. 

(c)  Current  required  to  operate  an  open-air  arc  lamp. 

(d)  Current  required  to  operate  a  2-motor  trolley  car  full  load. 

(e)  Current  required  to  operate  a  fan  motor. 

(/)   Current  required  to  operate  average  electric  bell. 

(g)  Current  required  to  operate  telephone  circuit. 

(h)  Current  required  to  operate  telegraph  circuit. 

(i)   Current   required   to    operate    150-volt  Weston  voltmeter 

for  full-scale  deflection. 
(j)   Current  required  to  operate  electric  welding. 


THE  FORMULATION  OF  PROBLEMS  135 

(k)  Current  required  to  operate  search  light. 

(I)    Current  required  to  operate  electric  locomotive. 

(m)  Current  required  to  operate  wireless  telegraph  outfit. 

(n)  Current  required  to  operate  railroad  signals,  marine  and 
hotel  annunciators. 

Ex.  12.  How  many  amp. -hours  will  be  recorded  by  a  meter 
through  which  150  amps,  have  passed  for  f  hr.? 

Ex.  13.  A  lOO-amp.-hour  Edison  cell  is  discharged  through 
an  electromagnet  at  a  3.5  amp.  rate.  How  long  will  the  cell 
maintain  this  current  through  the  magnet? 

Ex.  14.  A  meter  records  700  amp. -hours.  It  was  in  circuit 
7  days  for  10  hours  each  day.  What  was  the  average  rate  of 
current  used? 

Ex.  15.  How  many  coulombs  have  passed  through  an  arc 
lamp  in  f  hr.  if  the  current  is  9.6  amps.? 

Ex.  16.  What  current  strength  is  required  to  deposit  4  grams 
of  Cu  upon  an  iron  spoon  in  30  min. 

Ex.  17.  A  meter  records  44000  coulombs  in  8  hrs.  What  was 
the  average  strength  of  current? 

Ex.  18.  How  many  grams  of  Cu  will  be  deposited  on  an 
iron  plate  used  for  a  ship's  hull  in  10  hrs.  if  the  average  strength 
of  current  is  30  amps.? 

Ex.  19.  What  quantity  of  electricity  in  coulombs  passes  in  a 
circuit  in  which  a  current  of  40  amps,  flows  for  56  sees.? 

Ex.  20.  What  quantity  of  electricity  in  coulombs  passes  in 
a  circuit  of  13  amps,  flowing  for  15  min.? 

Ex.  21.  In  1  hour  36000  coulombs  of  electricity  pass  through 
a  closed  circuit.  If  the  flow  is  uniform  during  that  time,  what 
is  the  strength  of  the  current? 

Ex.  22.  How  long  will  it  take  72000  coulombs  of  electricity 
to  pass  in  a  circuit  in  which  the  strength  of  current  is  4  amps.? 


CHAPTER  V 

VARIATION   PROBLEMS   APPLIED   TO   THE   ELECTRIC 
CIRCUIT 

The  most  significant  fact  regarding  Ohm's  Law  is  that 
the  resistance  of  a  conductor  is  independent  of  the  strength 
of  the  current  flowing  through  it.  The  resistance  of  a 
conductor  depends  upon  its  shape,  size,  the  materials  of 
which  it  is  made,  and  also  upon  its  temperature  and  strain. 

The  resistance  of  a  conductor  varies  directly  as  its  length. 
This  fact  is  symbolized  in  three  different  ways : 

(1)     RczL,        (2)  j^-g        (3)     R  =  KL. 

Notation  : 
R  =  the  resistance  of  a  conductor, 
L  =  the  length  of  a  conductor, 
K\  =  a  proportionality  factor. 

Ex.  1.  2000  ft.  of  copper  (Cu)  wire  0.1  ins.  in  diameter  (dia.) 
has  2  ohms  resistance.  What  is  the  resistance  of  5000  ft.  of  the 
same  kind  of  wire? 

Ri    Lx  Data: 

{l)Rrz;  fl2=2u, 

fcJOOO  L2  =2000  ft, 

.  '    2      2000'  Rx=? 

(3)  #i=5,  Li  =5000  ft. 

Ex.  2.    Use  the  data  in  Ex.  1  and  determine:  the  resistances 
for  the  following  amounts  of  the  same  kind  of  wire: 

(a)  1000  ft.;  (6)  10000  ft.;  (c)  1  mile;  (d)  50  miles. 

136 


PEOBLEMS  APPLIED  TO  THE  ELECTRIC  CIRCUIT     137 

The  resistance  of  a  conductor  varies  inversely  (reciprocally) 
as  its  cross-section  area.     This  fact  is  symbolized  by: 

1  Notation: 

A  '  A  =the  area  of  the  section, 

Rx  _A2  K2=&  proportionality  factor, 

21  d  =the  dimensions  of  a  round  wire, 

(3)  #  =—-,  s=the  side  of  a  square  wire. 

A. 

t       xu  r  .Ax      Si2       __  ...     Rx      S22 

In  the  case  of  square  wires  —  = — .    Why?     /.     (4)  —  = — . 

A2    s22  J  w  R2    Si2 

Why? 

In  the  case  of  round  wires  —■»—.    Why?     .*.     (5)  -zr=~. 

A2    d22  J  w  R2    rfi2 

Why? 

Ax    ? 
In  the  case  of  conductors  of  similar  cross-section  —  =—.  W^hy? 

Jx2       i 

.'.     Rozj2    and    Roz^.    Why? 

Ex.  3.    Two  wires  of  circular  cross-section  have  resistances 
in  the  ratio  1:2.    What  is  the  ratio  of  their  diameters? 
Substituting  in  (5)  we  obtain: 

d22     1  /d2\*     1         .     dt       /I      1  dx      V2 

ds=2-  •••  U;  =2  and  71=\2=^-  •'•  sfTi--  Wh^? 

Ex.  4.  What  is  the  ratio  of  the  diameters  of  round  conductors 
whose  resistances  have  the  ratios:  (a)  1:3  (b)  1:4,  (c)  1  :  5, 
(d)  1  :  9,  (e)  1  :  16? 

Ex.  5.  What  is  the  ratio  of  resistances  of  round  conductors 
whose  diameters  have  the  ratios:  (a)  1:2,  (6)  1  :  3,  (c)  1  :  4, 
(J)  1  :  V2,  (c)  1  :  V3? 

Ex.  6.  Two  conductors  of  square  cross-section  have  resistances 
in  the  ratio  of  1  :  2.    What  is  the  ratio  of  their  dimensions? 

Using  formula  (4)  above  we  obtain 


SI.!.     ...     S-1=^L    and     s-l=^l 
si2    2  Si     V2  s2        1 


=    and  Why? 


138  PRACTICAL  MATHEMATICS 

Ex.  7.  What  is  the  ratio  of  dimensions  of  square  conductors 
whose  resistances  have  the  ratios :  (a)  1  :  3,  (6)  1:4,  (c)  1  :  5, 
(d)  1  :  9,  (e)  1  :  16? 

Ex.  8.  What  is  the  ratio  of  the  resistances  of  square  con- 
ductors whose  dimensions  have  the  ratios:  (a)  1  :  2,  (b)  1  :  3, 
(c)  1  :  4,  (d)  1  :  V2,  (e)  1  :  ^3? 

Ex.  9.  What  is  the  ratio  of  the  dimensions  of  two  conductors 
one  of  circular  and  one  of  square  cross-section,  which  have  equal 
areas? 

Ex.  10.  Two  conductors  are  equal  in  length  and  in  cross-section. 
One  is  a  round  wire  and  the  other  a  square  wire.  What  is  the 
ratio  of  their  radiating  surfaces? 

Observation.  The  resistance  of  any  round  conductor 
varies  jointly,  i.e.,  collectively  or  together  as  the  ength  and 
inversely  as  the  square  of  its  diameter. 

Reck 

from  which  we  obtain: 

(6)  «  =  -^- 


Notation:   • 

R  =  the  resistance  of  the  conductor  in  J2, 

L  =  the  length  of  the  conductor  in  ft., 

d  =  the  diameter  of  the  conductor  in  mils, 

K  =  a  proportionality  factor. 

The  value  of  K  may  be  determined  from  (6)  by  sub- 
stituting L  =  l  and  d  =  l  and  then  K  =  R,  i.e.,  K  equals 
the  resistance  of  1  ft.  of  wire  which  is  1  mil  in  diameter  and 
hence  K  is  called  the  circular  mil-foot  constant  or  mil-foot 
resistance. 

lmil  =  .001"=T^. 


PROBLEMS  APPLIED  TO  THE  ELECTRIC  CIRCUIT    139 

Table  of  Mil-foot  Resistances  at  0°C 

Silver 89.4        Iron 63.33 

Copper 9.35  Ger.  silver  (30%  nickel) .  .  290 

Aluminum 17  21      German  silver 128.29 

Manganin 258 .           Platinoid 188. 93 

Zinc 34.  69      Mercury 586.24 

Platinum 145 .  Nickelin  (40%  nickel). .  . .  290 

Ex.  11.    Determine  the  res.  of  1000  ft.  of  Cu  wire  TV  inch  dia. 
using  K  =  10.39. 

res.mil.ft.  xlength       KL       KL 


Resistance 


(diameter)2  (mils)2    CM' 

KL 


(7)  R     d2, 

10.39X1000 
H~        1002        ' 

fl  =  1.039. 

If  we  examine  a  wire  table  we  note  that  the  standard 
sizes  of  wires  are  numbered  and  the  nearest  size  to  a  100-mil 
wire  is  No.  10  B.  &  S.,  which  has  a  diameter  of  101.9  mils. 

Some  of  the  properties  of  the  wire  table  are  easily 
remembered  by  mnemonics  of  resistance.  A  mnemonic  is 
any  memory  device  which  suggests  the  association  of  ideas 
or  their  relations. 

Mnemonics  of  Resistance.  The  following  approximations 
serve  the  electrical  man: 

1  ohm  =  1000  ft.  of  Cu  wire  T\~  inch  dia.  (No.  10  B.  &  S.) 

1  ohm  =  250  ft.  of  Cu  wire  2\  inch  dia.  (No.  ?). 

Examine  a  wire  table  and  fill  in  the  blanks. 

1  ohm  =2  lbs.  of  Cu  wire  -£$  inch  dia.  (No ?). 

1000  ft,  of  Cu  wire  nearly  J§  inch  dia.  (0000  B.  &  S.)  =0.05 
ohms  approximately. 

1000  ft.  of  Cu  wire  y^  inch  (No.  40  B.  &  S.)  =  1063  ohms. 

The  area  of  any  cross-section  or  plane  figure  is  the 
ratio  of  the  figure  or  section  to  any  unit  of  area.  The 
common  unit  of  area  is  a  square  whose  sides  is  one  linear 


140 


PRACTICAL  MATHEMATICS 


unit.  If  the  side  of  the  unit  square  is  1  inch  then  the  unit 
of  area  is  1  sq.  in.  If  the  side  of  the  unit  square  is  1  mil 
then  the  unit  area  is  1  sq.  mil. 

In  wire  measure  the  unit  of  area  is  a  circle  having  a 
diameter  of  one  mil  and  therefore  the  unit  of  area  is  one 
circular  mil  (CM).  The  ratio  of  any  section  or  plane  fig. 
to  the  unit  CM  gives  the  area  in  CM,  whereas  the  ratio  of 
any  section  or  plane  figure  to  the  unit  sq.  mil  gives  the 
area  in  sq.  mils. 

Ex.  12.  How  many  circular  mils  in  a  wire  (a)  2  mils  dia., 
(b)  5  mils  dia.,  (c)  100  mils  dia.? 

Given  the  dia.  of  a  wire,  how  are  the  circular  mils  computed? 
Ex.  13.    What  is  the  cir.  mil  area  of  a  wire  \  inch  in  diameter? 


Fig.  20. 


Ex.  14.  Construct  a  square  three  mils  on  a  side  to  some  suit- 
able scale.  Divide  the  square  into  nine  smaller  squares  each  one 
mil  on  a  side.  Inscribe  a  circle  in  the  large  square  as  well  as 
in  each  of  the  smaller  squares.  Show  (a)  that  the  sum  of  the 
areas  of  the  smaller  circles  equals  the  area  of  the  larger  one; 
(b)  the  No.  cir.  mils  in  the  larger  circle  equals  9  times  the  No.  cir. 
mils  in  the  smaller;  (c)  the  No.  of  cir.  mils  in  the  cir.  cross-sections 
equals  the  No.  of  sq.  mils  in  the  square  cross-sections. 


PROBLEMS  APPLIED  TO  THE  ELECTRIC  CIRCUIT    141 

Ex.  15.  Show  that  the  cir.  mil  area  of  round  wires  is  nearly 
one-quarter  (0.2146)  larger  than  the  area  expressed  in  sq.  mils. 

Ex.  16.  A  wire  of  6530  CM  has  what  diameter?  Verify  by 
comparison  with  a  wire  table. 

Ex.  17.  A  copper  ribbon  for  a  field  coil  measures  |X|  ins. 
What  is  its  sq.  mil  area? 

Ex.  18.  What  is  the  equivalent  cir.  mil  area  for  the  cross- 
section  in  (17)? 

Ex.  19.    What  is  the  sq.  mil  area  of  a  wire  \  ins.  dia.? 

Ex.  20.  A  Cu  wire  has  a  cross-section  area  of  8234  CM  and 
has  a  length  of  1050  ft.    What  is  its  resistance? 

Ex.  21.  A  spool  is  to  be  wound  with  No.  20  B.  &  S.  German 
silver  wire.     What  length  is  required  to  give  50012? 

Ex.  22.  1000  ft.  of  iron  wire  has  a  resistance  of  30  12. 
What  is  its  CM  area  and  its  approximate  size? 

Ex.  23.  10  ft.  of  Cu  wire  has  a  res.  =.01312.  What  is  the 
gauge  number  and  the  resistance  of  1  mile  of  the  wire? 

Ex.  24.  A  German  silver  wire  of  11  ins.  has  0.022  12  resistance. 
What  length  is  required  to  give  2.412? 

Ex.  25.  1  mile  of  wire  has  a  resistance  of  14.7512.  What  is 
its  resistance  per  foot? 

Ex.  26.  The  resistance  of  18  ins.  of  wire  =0.02712.  What  is 
the  resistance  of  1020  ft.? 

Ex.  27.  The  res.  of  a  conductor  is  0.32  ohm  and  sectional 
area  =0.025  sq.in.  What  will  be  the  res.  of  a  like  conductor 
whose  sectional  area  =0.125  sq.in.,  other  conditions  being  the 
same? 

Ex.  28.  The  sectional  area  of  a  conductor  is  0.01  sq.in.  and 
its  res.  =1  ohm.  It  is  replaced  by  a  wire  0.001  sq.in.,  other  con- 
ditions being  alike.    What  will  be  the  res.  of  the  latter? 

Ex.  29.  A  round  copper  wire  0.12  in.  diameter  has  a  resist- 
ance 0.64  ohm,  whereas  a  round  copper  wire  0.24  in.  diameter  has 
a  resistance  ?  ohms,  when  other  conditions  are  alike? 

Ex.  30.  The  diameter  of  a  round  wire  is  0.1  in.  and  its  resist- 
ance =2  ohms,  whereas  the  diameter  of  a  round  wire  ?  in.,  and 
its  resistance  50  ohms,  when  other  conditions  are  alike? 

Ex.  31.  1000  ft.  of  copper  wire,  diameter  =0.05  in.,  has  a 
resistance  =4  ohms.  2500  ft.  of  copper  ribbon  0.006  in.  thick, 
0.02  in.  wide,  has  what  resistance? 

Ex.  32.  The  resistance  of  a  mil-foot  of  copper  wire  is  10.8 
ohms.  Using  the  same  quality  of  copper  determine  the  resistance 
for  the  following  wires  and  the  nearest  gauge  number  in  the 
wire  table: 


142  PRACTICAL  MATHEMATICS 

1200  ft.  of  wire  0.102  in.  dia. 

1  mile  of  wire  |  in.  dia. 

1500  ft.  of  square  wire  0.1  on  a  side. 

100  yds.  of  wire  0.12  wide  by  0.09  thick. 

Variation  Length,  CM  and  Weight 

lbs.  per  mile  bare  copper  wire  =  7^7-=, 

0J.0 

lbs.  per  foot  bare  copper  wire  = 


lbs.  per  foot  bare  iron  wire 


62.5X5280' 

CM 

72.13X5280* 


Ex.  33.    How  many  lbs.  of  No.  10  Cu  wire  are  required  if  strung 
5  miles  and  return? 

Ex.  34.     Give    the    equivalent    resistance    in    microhms    of 
0.00425  ohm. 
1  microhm  =  .00000112,  1  milliohm  =  .0010,  1  megohm  =  lOOOOOOfi. 

Give  the  equivalent  resistance  in  ohms  of  375  microhms. 

Give  the  equivalent  resistance  in  mehgoms  of  4560000  ohms. 

Give  the  equivalent  resistance  in  ohms  of  62.5  megohms. 

Give  the  equivalent  resistance  in  milliohms  of  5  megohms. 

Use  the  abbreviated  notation  1000000  - 106. 

Ex.  35.    The  insulation  res.  of  a  wire  measures  16.75  megohms. 
What  is  its  equivalent  res.  in  ohms? 

Ex.  36.    What  is  the  CM  area  of  a  wire  ^  in.  in  diameter? 

Ex.  37.    The  CM  area  of  a  wire  is  5625.    What  is  its  dia.? 

Ex.  38.    An  armature  is  wound  with  copper  bars  ft  Xf  of  an 
inch.    What  is  their  equivalent  area  in  cir.  mils.? 

Ex.  39.    The  res.  of  a  series  coil  of  a  dynamo  is  0.0065  ohm. 
What  is  the  equivalent  in  microhms? 

Ex.  40.    What  is  the  sq.  mil  area  of  a  No.  12  B.  &  S.  copper 
wire? 

Ex.  41.    What  is  the  res.  of  5  lbs.  of  No.  18  B.  &  S.  wire 
allowing  5  per  cent  for  insulation? 

Ex.  42.    The  coils  of  a  rheostat  are  constructed  of  No.  8  iron 
wire  and  have  a  res.  of  10  ohms.    What  length  of  wire  was  required? 

Ex.  43.    A  rectangular  conductor  has  a  sq.  mil  area  of  20616.75. 
What  is  its  equivalent  CM  area?  1 

Ex.  44.    What  size  of  B.  &  S.  wire  has  an  area  equivalent  to 
the  wire  in  the  preceding  problem? 


PROBLEMS  APPLIED  TO  THE  ELECTRIC  CIRCUIT     143 

Ex.  45.  Calculate  the  res.  of  2000  ft.  of  No.  6  B.  &  S.  copper 
wire. 

Ex.  46.  Construct  from  your  own  calculations  by  the  use  of 
formulas  a  wire  gauge  table  for  No.  12  B.  &  S.  copper  wire. 

Give  cir.  mil  area,  sq.  mil  area,  lbs.  per  mile,  lbs.  per  ft.,  lbs. 
per  ohm,  ft.  per  lb.,  ft.  per  ohm,  ohms  per  lb.,  and  ohms  per  ft. 
A  No.  12  B.  &  S.  wire  has  a  diameter  =  .08081  in. 

Ex.  47.  What  is  the  area  in  CM  of  a  wire  having  a  diameter 
of  0.46  in.? 

Ex.  48.  Find  the  area  of  a  copper  rod  having  a  diameter 
of  A  in. 

Ex.  49.  What  is  the  diameter  of  a  wire  having  a  sectional 
area  of  1021.5  CM? 

Ex.  50.  Brown  &  Sharpe  or  American  Gauge.  Show  that  a 
No.  7  wire  has  a  sectional  area  equal  to  two  No.  10  wires. 

Show  that  four  No.  13  wires  have  a  sectional  area  equal  to 
eight  No.  16  wires. 

Ex.  51.  Show  that  by  subtracting  3  from  any  gauge  number 
we  obtain  the  number  of  a  wire  having  very  nearly  twice  the 
sectional  area.  Compare  Nos.  1,  2,  ...  10  with  Nos.  10, 
11  .  .  .  20. 

Ex.  52.  Show  that  the  ratio  between  the  resistance  of  any 
wire  in  the  B.  &  S.  gauge  and  that  of  the  next  higher  number  is 
that  of  1  :  1.26,  i.e.,  1  :  i/2. 

Ex.  53.  How  should  the  above  statement  read  when  the 
ratio  is  inverted? 

Ex.  54.  What  is  the  res.  of  1000  ft.  No.  16  wire  having  given 
the  res.  of  1000  ft.  No.  10  equals  1  ohm? 

Ex.  55.  The  res.  of  a  No.  12  B.  &  S.  gauge  copper  wire  is  8.37 
ohms  per  mile.  What  is  the  res.  of  a  No.  11  and  also  a  No.  13 
wire? 

Ex.  56.'  The  res.  of  a  No.  00  copper  conductor  is  0.411  ohm 
per  mile.    What  is  the  res.  of  a  similar  conductor  of  No.  3  gauge? 

Ex.  57.  It  is  a  very  convenient  fact  to  remember  that  the 
diameter  of  a  No.  10  wire  in  the  B.  &  S.  gauge  is  very  close  to  Tlo 
in.  (0.10189)  and  that  the  res.  of  a  No.  10  annealed  copper  wire 
per  1000  ft.  is  practically  1  ohm  (0.9972).  Determine  the  per 
cent  of  error  for  the  two  approximate  values. 

Ex.  58.  Determine  the  sizes  of  wire  for  supplying  200  lamps 
at  110  volts  with  an  allowable  loss  of  10  per  cent  in  the  mains, 
when  two-  and  three-  wire  system  are  used.  Calculate  the  difference 
in  copper  allowing  in  the  3-wire  system  a  neutral  wire  of  half 
the  cross-section  of  the  outer  wires. 


144 


PRACTICAL  MATHEMATICS 


Make  the  necessary  drawings  to  illustrate  the  problems,  and 
calculate  the  percentage  on  the  power  delivered  to  the  receiving 
circuit.    Fig.  21. 


<5 


.055  Ohms 


.055  Ohms 


100  Amps. 


100  Ampe. 


75777 


200 
Lamps 


t 


50  Amps. 


a  is 

&    I 
50  Amps.         -f   y 


Fig.  21. 


Ex.  59.  The  data  are  the  same  as  in  Ex.  58,  except  the  loss 
in  the  mains  is  5  per  cent. 

Ex.  60.  An  armature  of  an  alternator  has  a  diameter  of  4.5 
ft.  and  runs  at  300  revolutions  per  minute.  What  is  its  peripheral 
speed? 


CHAPTER  VI 
THE   ALGEBRA   OF   A   SIMPLE   ELECTRIC   CIRCUIT 

The  notation  used  in  this  chapter  is  given  below. 

/  =  current  amperes;   also  *i,  ^2,  etc.,  instantaneous 

values  current, 
#  =  E.M.F    or  P.D.  or  drop  in  volts;    also  e\,  62, 

etc.,  instantaneous  values  E.M.F., 
R  =  resistance  in  ft;  also  X,  M,  N,  Ri,  R2,  m,  n,  n9 
T2,  ?%  r±, 
Er,  Ex  =  drop  in  volts   corresponding   to  res.  R  and  X 
respectively, 
X  =  unknown  resistance, 
.Rr  =  res.  at  temp.  T, 
Ro  =  Tes.  at  temp,  zero  centigrade, 
K  =  res.  of  a  mil-foot  wire, 
L  =  length  of  wire, 
A  =  area  of  cross-section, 
K\j  K2  =  constants  of  instruments, 
a  =  angle  of  deflection, 
W  =  watts, 
s  =  No.  cells  in  series, 
p  =  No.  cells  in  parallel, 
H.P.  =  horse-power, 
K.W.  =  kilowatt, 

T  =  temperature  in  centigrade  degrees. 

The  subscripts  are  often  replaced  by  accents  as  shown 
in  Ex.  8.  In  Ex.  11  sta  and  mov  are  subscripts  which 
indicate    the    respective    currents    in    the    stationary    and 

145 


146  PEACTICAL  MATHEMATICS 

movable  coils.     In  formula  work   multiplication    symbols 
are  avoided  so  as  not  to  conflict  with  X  in  the  notation. 

Transform  and  interpret  the  following  equations  for  each  letter 
and  solve  for  the  numeric  values  when  data  are  given  in  the 
following : 

Ex.  1.     Measurement  of  resistance: 

RI=E. 
Solve  for  R  and  /. 

Ex.  2.    Drop  of  potential  in  a  conductor: 

Er    Ex 
R  ~  X' 

Solve  for  ER,  R,  Ex,  X,  giving  all  possible  values. 

Ex.  3.  Influence  of  length,  cross-section  and  material  of  a 
conductor  on  its  resistance : 

A 

Solve  for  K,  L,  A. 

Ex.  4.  Influence  of  temperature  on  the  resistance  of  con- 
ductors: 

(1)  ftr  =#o(l+0.0042  T). 

If  the  resistance  R0  at  0°  centigrade  be  taken  as  100  per  cent, 
what  is  the  per  cent  increase  of  temperature  per  each  degree 
centigrade? 

For  commercial  Cu  the  formula  is  written 

(2)  flr  =#o(l+0.004  T). 

Determine  the  R  at  25°  C.  and  15°  C.  in  terms  of  Rq  and  show 
the  per  cent  error  in  using  (2)  instead  of  (1). 

Ex.  5.  In  the  preceding  work  eliminate  Rq  by  dividing  the 
equation  for  RT  by  the  equation  for  Rn. 

Determine  T  given  RT  =5.468  #15  =4.573. 

Ex.  6.    Resistances  in  parallel: 


i  J-    i'JL 

Ill  tit 


Solve  for  -1. 

i2 


THE  ALGEBRA  OF  A  SIMPLE  ELECTRIC  CIRCUIT  147 


The  elements  of  a  branch  are  marked  with  like  subscripts. 
Ex.  7.    Wheatstone   bridge.    When  no  current  flows  through 
the  galvanometer,  i.e.,  a  balance,  then 


(1) 
(2) 


Ri1=Mi2=E1. 

Xil=Ni2=E2. 

Divide  (1)  by  (2)  and  solve  for  X. 
X  is  the  arm  of  unknown  resistance. 

Data: 
X  =  l,  #  =  10  ohms, 

M  -  27  ohms,    N  =  73  ohms. 


Fig.  22. 

Ex.  8.     Measurement  of  very  small  resistances  by  the  Kelvin 
double  bridge :    Given 


m  _m 
n      n' 


(1) 

and 

(2) 

then 

(3) 


by  the  Ax  of  fractions  and  =tyAx. 
But  we  have  also  given 

(4)  i'm'=IR+im, 
and 

(5)  t'n'«JX+tn; 


R     m 

~X=n' 

R 
X 

IR 
~IX 

m    mi 

n     ni 

m' 

~n'  '' 

m'i' 
~  n'i' ' 

148  PRACTICAL  MATHEMATICS 

show  that 

R     i'm' —im 


X     i'n'  —in 
Ex.  9.    Soft  iron  instrument : 

¥u\\=IK2) 

?U\\=KJ*. 

Solve  for  Kh  K2,  I,  eliminating  pull. 
Ex.  10.     Ammeter  shunt : 

I=E  Data: 

R'  #=0.050  volt, 

72=0.001  ohm, 

/=?. 

Ex.  11.     Electro-dynamometer  instrument: 


(a)  torque  = 

(b)  tgta  = 

(c)  torque  = 

(d)  torque  = 

-^M^sta^movj 

=Ki  times?, 

»ve 

(«)           i- 

(/)            i- 

fir 

What  is  the  relation  between  i£3,  -K^,  £i,  if  Eq.  (/)  is  made 
to  replace  (e)?. 

Observation.  The  ratio  or  product  of  two  constants  is 
another  constant.  The  root  or  power  of  a  constant  is  another 
constant 

Ex.  12.     Standard  cell: 

E  =  1.0186  -O.OOOOSSCT7 -20°)  -  0.00000065  (T7 -20°)2. 

Compute  E  for  the  following  temperatures:  (a)  10°,  (b)  15°, 
(c)  20°,  (d)  25°,  (e)  30°. 


THE  ALGEBRA  OF  A  SIMPLE  ELECTRIC  CIRCUIT  149 
Ex.  13.     Indicating  wattmeters : 

Watts  =  amperes  X  volts, 
W=IE. 
Solve  for  /  and  E  and  state  the  formula  as  a  law. 

Ex.  14.     Compensation  for  power  consumption  in  the  watt- 
meter: 

Solve  for  i,  R,  E. 

Ex.  15.     Circuit  with  series  resistance: 

E 


R+n+rt 


Solve  for  E,  R,  rh  r2,  n  +r2. 
Ex.  16.     Resistances  in  parallel: 

11111 

a     n     r2    rz    r4 

Solve  for  R,  rh  r2,  rZj  n. 

Interpret  this  formula,  naming  the  reciprocal  of  a  resistance 
as  conductivity.  It  will  be  noticed  that  in  any  of  the  trans- 
formations of  this  example  the  small  r's  are  homogeneous,  i.e., 
interchangeable. 

Ex.  17.     Cells  in  series: 

.  _    Es  Notation: 

rs+R'  r  =  resistance  of  a  cell, 


Solve  for  E,  s,  r,  R. 

Ex.  18.     Cells  in  parallel: 


/  =  - 

r 


R  =  external  resistance. 


E 


+R 
V 


Solve  for  E,  r,  p,  R. 


150  PRACTICAL  MATHEMATICS 

Ex.  19.     Cells  in  mixed  combination: 

V 
Solve  for  E,  s,  r,  p,  R. 
Ex.  20.     Formulas  for  wattage: 

'-# 

In  (a)  sub.  for  (7)  from  (1)  and  solve  for  W  and  E. 
Ex.  21.     Mechanical  horse-power  of  electric  circuit. 

«  ttP"f§ 

In  (a)  sub.  for  7  from  (1)  and  solve  for  E  and  72. 
In  (a)  sub.  for  #  from  (1)  and  solve  for  E  and  7. 


CHAPTER  VII 
THE   APPLICATIONS   OF   OHM'S    LAW 

The  current  flowing  through  an  electric  circuit  is  subject 
to  change  arising  from  a  variation  in  the.value  of  its  resistance 
or  in  the  value  of  the  impressed  E.M.F. 

Ohm's  law  contains  the  three  elements  7,  E,  and  R. 
Write  its  three  forms.  Consider  R  a  constant,  i.e.,  a 
fixed  resistance;  (a)  how  can  E  and  I  be  made  to  change; 
(b)  what  is  the  effect  upon  either  E  or  I  if  one  of  them 
increases;  (c)  if  one  of  them  decreases;  (d)  if  E  vanishes 
what  is  the  effect  upon  the  current;  (e)  if  I  is  very  excessive; 
what  is  the  corresponding  effect  upon  the  value  of  E;  (/) 
does  a  direct  or  indirect  variation  exist  between  E  and  I? 
(g)  write  the  variation,  using  the  variation  symbol;  (h)  what 
is  R  in  such  a  variation? 

In  the  three  forms  of  Ohm's  law  substitute  the  value  of 
R  in  terms  of  length,  cross-section,  and  temperature. 

Restate  all  the  possibilities  for  changing  the  current 
in  a  circuit  when  the  impressed  E.M.F.  remains  constant 
but  when  the  elements  of  resistance  are  made  to  change. 

Determine  the  numeric  values  in  the  following  examples: 
Ex.  1.  Data: 

t    £*  Incandescent  lamp. 

R'  #  =  110  volts, 

R  =200  ohms, 

Ex.  2.  Data: 

t  _E_  Arc  lamp. 

R'  E  =  50, 

R=?, 
1=7. 

151 


152  PRACTICAL  MATHEMATICS 

Ex.  3. 


Ex.  4. 


Data: 

<-§■ 

Absolute  units. 
E  =  10\ 
R  =  10\ 
/  =  ?. 

Data: 

7=   2  . 

R+r 

Battery  circuit. 

R  =  60  external  resistance, 
r=3  internal  resistance. 

1  =  2  amperes. 

Ex.  5.  An  incandescent  lamp  has  a  hot  res.  220  ohms  and 
is  connected  to  an  electric  light  main  across  which  110  volts 
potential  difference  is  maintained. 

Ex.  6.  A  circuit  has  a  resistance  of  50  ohms  and  a  pressure 
of  110  volts.     What  is  the  strength  of  the  current  in  amperes? 

Ex.  7.  The  pressure  in  a  conductor  is  4  volts  and  the  resistance 
15  ohms.     How  many  amperes  is  flowing? 

Ex.  8.  What  current  can  be  made  to  flow  through  a  circuit 
having  a  resistance  of  10  ohms,  if  an  E.M.F.  of  110  volts  is 
applied? 

A  resistor  is  any  coil,  conductor,  or  mechanical  device 
whose  resistance  is  used  specifically  for  controlling  the 
strength  of  a  current  of  electricity. 

Ex.  9.  A  resistor  is  connected  to  a  550- volt  circuit.  What 
resistance  must  it  have  in  order  that  a  current  of  .5  ampere 
may  flow  through  it? 

E 

By  Ohm's  law  R=j  and  after  substituting  the  values  of  E 

and  I  from  the  data,  we  obtain: 

Data: 

B-^-UOOQ.  '  'f  ydto- 

.5  1  =.5  amp. 

Ex.  10.  What  is  the  resistance  of  a  resistor  in  order  that  a  cur- 
rent of  50  amperes  may  flow  through  it  if  it  is  connected  to 
500-volt  mains? 


THE  APPLICATIONS  OF  OHM'S  LAW  153 

Drop  of  Potential.  The  E.M.F.  between  two  points  of 
a  circuit  is  variously  designated  as  a  potential  difference 
(P.D.),  or  pressure  difference  in  volts  or  millivolts,  and 
more  commonly  as  the  drop,  meaning  the  fall  or  difference 
of  voltage. 

Ex.  11.  Fig.  23  represents  a  part  of  a  circuit,  A,  B,  C,  D,  E 
being  definite  points  thereon.  A  current  of  2.5  amperes  flows 
through  the  conductor.  The  drops  between  the  two  points  desig- 
nated are  given  as  follows: 

45  =  12.5  volts,  5(7  =  6.25  volts,  CD  =  18  volts,  DE  =  .025  volt. 

The  resistances  corresponding  to  these  drops  are  computed  by 

E 

substituting  the  data  in  Ohm's  law:  R=y.    Therefore  AB=5tt, 

£C  =2.512,  CD  =7 .20,,  DE  =  .0112. 


Fig.  23. 

Ex.  12.  It  is  desired  to  transmit  10  amperes  to  supply  power 
1000  ft.  from  the  source  at  110  volts.  What  size  conductor  would 
be  required  for  a  2  per  cent  loss  in  the  transmission? 

110 X. 02  =2.2  volts  drop  in  the  line, 

2  2 

—=1.1  volts  drop  in  each  main, 
2  F 

E     1.1 

.1112,  resistance  of  each  main, 


ft 

/ 

Tu=-in''re! 

KL 

R 

~  d*' 

tP 

KL 

10.8X1000 

R 

.11 

d2  =98182  CM  in  the  cross-section. 

From  the  wire  table  we  find  No.  0  wire  is  required. 
Ex.  13.     The  millivolt  drop  across  3  ft.  of  rail  including  a  bond 
is  20  and  the  corresponding  drop  for  3  ft.  of  continuous  rail  is 


154  PEACTICAL  MATHEMATICS 

15.     Taking  these  as  average  values  what  is  the  estimated  drop 
per  mile  of  track  (using  30  ft.  rails)? 

Ex.  14.  In  a  closed  circuit  the  drop  caused  by  the  resistance 
of  the  conductor  is  10  volts.  If  the  current  flowing  is  4  amps, 
what  is  the  resistance  of  that  part  of  the  circuit? 

Ex.  15.  The  E.M.F.  generated  in  a  circuit  is  220  volts.  The 
current  is  10  amps.  The  leads  have  a  drop  of  10  per  cent.  What 
is  their  resistance,  (a)  when  per  cent  is  figured  on  generator 
E.M.F.,  (6)  when  per  cent  is  figured  on  E.M.F.  delivered  to 
receiving  circuit? 

Ex.  16.  How  much  pressure  will  it  take  to  force  a  current 
of  18  amps,  through  a  resistor  of  5  ohms? 

Ex.  17.  What  voltage  is  required  to  send  a  current  of  25 
amps,  through  a  resistor  of  4  ohms? 

Ex.  18.  The  total  resistance  of  a  closed  circuit  is  49.3  ohms. 
If  the  current  is  2.73  amperes,  what  is  the  total  E.M.F.  in  volts? 

Ex.  19.  A  difference  of  potential  of  110  volts  exists  between 
the  terminals  of  a  conductor  whose  resistance  is  20  ohms.  What 
is  the  current  flowing  through  the  conductor? 

Ex.  20.  A  circuit  has  an  available  pressure  of  220  volts. 
What  is  its  resistance  if  a  current  of  50  amps,  can  flow  through  it? 

Ex.  21.  The  two  electrodes  of  a  simple  voltaic  cell  are  connected 
together  by  a  copper  wire  having  1  ohm  resistance.  If  the  inter- 
nal resistance  of  the  cell  is  2.4  ohms  and  E.M.F.  2  volts,  what  is 
the  strength  of  current  in  the  circuit? 

Ex.  22.  The  total  E.M.F.  developed  in  a  circuit  is  1.2  volts 
and  the  strength  of  the  current  is  0.3  amp.  What  is  the  total 
resistance  of  the  circuit? 

Ex.  23.  A  battery  of  10  cells  each  having  an  E.M.F.  2  volts 
and  internal  resistance  2.412  causes  a  current  of  .25  amp.  in  a 
circuit.     What  is  the  external  resistance  of  the  circuit? 

Ex.  24.  Given  a  battery  circuit  in  which  the  E.M.F.  =30 
volts,  current  =  .6  amp.,  internal  resistance  =40  ohms,  what  is 
the  external  resistance? 

Ex.  25.  The  total  E.M.F.  of  a  voltaic  battery  is  22  volts 
and  its  total  internal  resistance  is  11  ohms.  What  is  the  external 
resistance  of  the  circuit  if  0.5  amp.  is  flowing  through  it? 

Ex.  26.  A  telegraph  circuit  has  two  sounders  of  20  ohms 
each  in  series,  the  line  resistance  is  20  ohms.  The  fine  is  operated 
by  two  sets  of  10  cells  each.  Each  cell  has  an  E.M.F.  of  1.5  volts 
and  a  resistance  of  3  ohms.  What  current  flows  through  the* 
circuit  (a)  when  the  two  sets  of  cells  are  in  series,  (6)  when  the 
two  sets  of  cells  are  in  parallel? 


THE  APPLICATIONS  OF  OHMS  LAW 


155 


Fig.  24  is  a  sketch  of  the  connections  showing  a  double  throw 
switch  which  permits  the  series  arrangement  in  position  a  and 
the  parallel  arrangement  in  position  b. 

R  =  The  total  battery  resistance +the  resistances  of  the  sounders 
and  line, 

E=the  total  battery  E.M.F. 

/=  current. 


ji  IK 


20  Ohms  each 


^zPD 


lay 

\  1.5  Volts!  „    , 
V  3  Ohms|  Each 

Fig.  24. 


Arrangement  (a): 

R  =  20X3  +40  +20  =  12012, 


arrangement  (b): 

R  =i^ +40+20=75, 


E  =20X1.5  =30  volts, 

T      E       30         O* 

/=S  =  l20=-2°ampS' 


#  =  10x1.5  =  15  volts, 

7      E      15       9 


Ex.  27.  How  many  electric  heaters  of  1012  each  can  be  con- 
nected in  series  to  a  trolley  pressure  of  550  volts  so  as  to  draw  5 
amperes? 

N  =  number  of  heaters, 

R  =  10  N  =  total  resistance, 

1  5 

.*.     N  =  11,  the  number  of  heaters  connected  in  series. 

Heater  Heater  Heater  Heater 

VvWvy\A/» tyMA/W^> — vwvww — 'wvwyw 


vSV  550  Volts 
5  Amps. 

Fig.  25. 

Ex.  28.  The  electric  heaters  on  a  street  car  draw  5  amperes 
at  a  trolley  pressure  of  550  volts.  There  are  four  heaters  in 
series,  Fig.  25.  What  is  the  resistance  of  each  heater?  How 
much  power  does  it  consume? 


156  PRACTICAL  MATHEMATICS 

Ex.  29.  The  hot  resistance  of  the  field  coil,  Fig.  26,  of  a  shunt 
dynamo  is  10  ohms.  How  much  current  is  it  drawing  when  the 
machine  is  generating  220  volts? 

j_E_220 
~R~W 

I  =22  amperes  passing  through  the  field  coil. 


10  Ohms  ^Armature  tmnv.m.  ^Armature 


20  Volts  S  Vp7 180  Volts 

5  Ohms 

Fig.  26.  Fig.  27. 

Ex.  30.  When  a  5-ohm  resistor,  Fig.  27,  is  inserted  in  series 
with  the  field  coil  of  the  above  machine,  the  armature  pressure 
drops  to  180  volts.     What  current  flows  through  the  field  coil? 

Ex.  31.  The  open-circuit  E.M.F.  of  a  railway  storage  battery, 
Fig.  28,  is  580  volts.  When  500  amps,  are  drawn  from  it  the 
E.M.F.  falls  to  540  volts.     What  is  the  resistance  of  the  battery? 

E2  (open  circuit)  =580, 

Ei  (closed  circuit)  =540, 

E=E2  —  El  =  40  =  volts  drop  in  the  battery 

^~/_500' 
R  =  .0812  resistance  of  the  battery 


500  Aid  pa. 

1^510  Volts^ 


Fig.  28. 

The  power  consumed  in  an  electric  circuit  is  the  product 
of  its  voltage  and  its  amperage  measured  simultaneously. 

watts  =  volts  X  amperes,  P  =  EI. 


THE  APPLICATIONS  OF  OHMS  LAW  157 

Ex.  32.     A  550- volt  trolley  circuit  is  supplying   100  amperes 
to  a  trolley  car,  Fig.,  29.     The   line   and  track  resistance  is  .512. 

100  Amps. 

550  Volts  fa  1,5  Ohms  '|qSd  f 

C_ J £  BE 

Fig.  29. 


What  is  the  pressure  available  at  the  car  and  what  power  does 
it  consume? 

E  =  /#  =  100x.5, 

E  =  50  volts  drop  in  the  line  and  track, 
E\  =550  -50  =500  volts  available  at  the  car, 
P  =  EJ  =500X100, 

P  =50  K.W.,  the  power  used  by  the  car. 

Ex.  33.  If  an  electric  motor  is  delivering  5  horse-power  at 
an  efficiency  of  80  per  cent,  what  current  is  it  drawing  from  a 
220-volt  line? 

Ex.  34.  A  residence  contains  25  incandescent  lamps,  Fig. 
30,  which  operate   on  an  average  of  20  hours  per  month.    Each 

no  voHs 


t  tettk- 


25  Lamps 


\mps. 

Fig.  30. 

lamp  takes  0.6  amp.  at  110  volts.     What  is  the  monthly  bill  at 
11  cents  per  K.W.  hour? 

/  =25 X.6  =  total  current  used, 

P=#/  =  110x25X.6  =  1650watts  =  1.65K.W. 

K.W.  hours  =  1.65X20  =33, 

33X.11  =$3.63,  monthly  bill. 


158  PKACTICAL  MATHEMATICS 

Ex.  35.     A  20-ton  trolley  car,  Fig.  31,  uses   0.18  K.W.  hour 
per  ton  mile.     What  is  the  average  current  with  a  trolley  line 


€ 


ffij 


Fig.  31. 

pressure  of  525  volts  when  the  car  speed  averages  15  miles  per 
hour? 

Data: 

.18  X 1000  =  180  watt  hours  per    weight  of  car  =20  tons, 

ton  mile,  capacity  =  .18  K.W.  hour 

180  X 15  X  20  =  watt     hours     con-  per  ton  mile, 

sumed.  speed  =  15    miles    per 

Complete  the  problem.  hour. 

Ex.  36.  A  railway  storage  battery  is  discharged  at  a  rate  of 
1100  amps,  for  eleven  hours.  How  many  ampere-hours  has  it 
given  out?    How  many  coulombs? 


Fig.  32. 

Ex.  37.  How  many  days  will  a  10-volt  gravity  battery,  Fig. 
32,  continue  to  yield  0.1  amp.  if  its  ultimate  output  is  10  K.W. 
hours? 

Data: 
watts  =#7  =  10X.1=1,  E  =  10,         7  =  .l, 

output  =  10  K.W.  hours. 

watt  hours 

hours  = — . 

watts 

Complete  the  problem. 


THE  APPLICATIONS  OF  OHM'S  LAW 


159 


Ex.  38.  A  resistance  frame,  i.e.,  a  subdivided  resistor,  Fig. 
33,  is  to  be  made  of  coils  of  wire  in  series  of  sufficient  size  to  carry 
10  amps.  How  must  the  resistance  be  divided  to  provide  taps 
in  gradations  of  0.5,  1,  2,  3,  5,  10,  20,  30,  50,  100,  150  volts? 


NWv\M 


F\V^vVVv\AAAAAhAA/%VvVV1 


Resistance  in  Ohms 


Fig.  33. 


When  a  number  of  resistors  are  connected  in  parallel 
the  reciprocal  of  their  combined  resistances  equals  the 
sum  of  the  reciprocals  of  the  individual  resistances, 

R    nTr2Tf8T 

Ex.  39.  A  current  of  .6  ampere  is  supplied  to  two  resistors, 
n  and  r2,  in  parallel,  Fig.  34.  What  current  flows  through  each 
resistor  when  n  =212,  and  r2  =  312? 

I-i+i. 

R    n    r2 

Ex.  40.  What  is  the  joint  resistance  of  n=412  and  r2=612 
when  joined  in  parallel? 


f 


T  •  .6  Amps.     Kf  '  | 1 


T 


Fig.  34. 


Fig.  35. 


Ex.  41.    What  is  the  joint  resistance,  Fig.  35.  of  three  parallel 
conductors  when  n  =512,  r2=  1012,  and  r3  =2012? 


160 


PRACTICAL  MATHEMATICS 


Ex.  42.  The  sum  of  the  currents  in  the  three  branches  rh 
r2,  r3,  of  a  divided  circuit  is  52  amps.  n=4,  r2  =  6,  r3=8  ohms. 
How  much  current  flows  through  each  branch? 

Ex.  43.  Four  resistors  A,  B,  C,  D,  are  connected  in  parallel. 
The  resistances  are  4,  5,  8,  10  ohms  respectively.    If  the  current 


r 

T 


;A       IB      |C 


Fig.  36. 

through  A  alone  is  40  amps.,  how  much  current  will  flow  through 
each  of  the  other  resistors? 

Ex.  44.     Two  resistors,  A  and  B,  are  in  parallel  and  joined 
to  a  resistor  C  in  series,  Fig.  37.    A  =40,  Z?=60,  (7=80  ohms. 


-nov. 
Fig.  37. 


A  voltage  of  110  is  connected  to  the  extremities  of  the  com- 
bination.   What  current  flows  through  each  resistor? 

Ex.  45.     A  generator,  Fig.  38,  supplies    5.96   amperes    to  a 
multiple-series  circuit.    The  brush  voltage  is  110  volts  and  the 


5.96  Amp.  ^C 

t        A        \ 

1  "\            HO  V. 

Fig.  38. 


resistance  of  the  generator  is  112.    What  voltage  will  be  shown 
at  the  brushes  on  open  circuit? 


THE  APPLICATIONS  OF  OHM'S  LAW 


161 


Ex.  46.  Two  copper  wires  1000  ft.  each  are  joined  in  parallel, 
Fig.  39.  One  is  a  round  wire  0.02  in.  diameter,  the  other  a 
square  wire  0.02  in.  on  a  side.  How  will  a  current  of  3  amps, 
distribute  itself  in  these  wires? 


27fi 


3  Amps. 


*  *1 


^2=21.2^ 


Fig.  39. 

Es.  47.  Determine  the  power  consumed  in  a  circuit,  Fig. 
40,  having  17.512  resistance  when  an  E.M.F.  of  110  volts  is 
applied  to  it? 

Ex.  48.  Compute  the  K.W.  consumed  in  a  circuit,  having  a 
resistance  =212  when  110  volts  is  applied  to  it? 


no  v. 


17.51) 


20  A 


11.8 


Fig.  40. 


Fig.  41. 


T 


112  V. 


X 


Fig.  42. 


Ex.  49.     Compute  the  kilowatts  consumed  in  a  circuit,  Fig. 

41,  having  a  resistance  =  11.812  and  a  current  =20  amps. 

Ex.  50.     Compute  the  kilowatts  consumed  in  a  circuit,  Fig. 

42,  having  an  E.M.F.  =  112  volts  and  a  current  =  12  amps. 

Ex.  51.     Galvanometer  shunts  are  resistors  which  are  adjusted 
to  i,  9V,  and  -9^9  of  the  resistance  of  the  galvanometer. 


Fig.  43. 


Write  a  formula  for  the  current  flowing  through  a  shunted 
galvanometer  circuit  coutaining  an  external  resistance.     Designate 


162  PRACTICAL  MATHEMATICS 

the  resistance  of  galvanometer,  shunt,  external  resistance  by  Rg, 
Rs,  and  Re,  respectively. 

Calculate  the  currents  in  the  main  circuit,  in  the  shunt,  and 
in  the  galvanometer  branches  when  the  battery  E.M.F.  =2  volts, 
Ra= 1000,22*  =50. 

Ex.  52.  What  is  the  rate  of  doing  work  in  watts  when  a  cur- 
rent of  50  amps,  flows  against  a  resistance  of  2  ohms? 

The  power  expended  in  a  circuit  is  directly  proportional 
to  the  square  of  the  voltage  impressed  upon  the  circuit  and 
inversely  proportional  to  the  resistance  of  the  circuit. 

Ex.  53.  How  many  watts  are  expended  in  a  resistor  in  over- 
coming its  resistance  of  220  ohms  when  a  pressure  of  110  volts  is 
applied  to  the  terminals  of  the  resistor? 

Ex.  54.     How  many  watts  are  equivalent  to  10  horse-power? 

Ex.  55.     How  many  K.W.  are  equivalent  to  5000  horse-power? 

Ex.  56.  The  power  in  a  circuit  is  equivalent  to  5  horse-power. 
What  current  will  flow  to  correspond  to  a  110  volt  pressure? 

Ex.  57.  Two  car-  motors  consume  75  amperes  at  550  volts. 
What  is  the  equivalent  horse-power? 

Ex.  58.  A  resistor  of  2  ohms  is  placed  in  series  with  two 
arc  lamps  so  that  a  current  of  9.6  amperes  flows  when  connected 
with  a  110- volt  circuit.    What  is  the  drop  across  each  lamp? 

In  making  resistors  of  small  resistance  it  is  convenient 
to  cut  a  piece  of  material  approximately  near  but  slightly 
in  excess  of  the  required  value  and  then  to  adjust  a  shunted 
wire  around  it. 

Ex.  59.  A  piece  of  wire  of  .10212  is  to  be  adjusted  to  make 
a  resistor  of  .112.  (a)  What  resistance  of  wire  is  required  to 
shunt  it?  (b)  Suppose  an  error  of  2  per  cent  is  made  in  the 
adjustment  of  the  shunt,  how  will  it  affect  the  desired  result? 

Ex.  60.  What  power  in  kilowatts  will  be  taken  by  a  10-horse 
power  motor  having  an  85  per  cent  efficiency?  At  a  pressure 
of  220  volts  what  amperage  is  required? 

The  power  expended  in  overcoming  the  resistance  of 
lamps,  resistors,  machines,  leads,  and  other  devices  in  a 
circuit  is  proportional  to  their  resistance  and  to  the  square 
of  the  current  passing  through  them. 


THE  APPLICATIONS  OF  OHM'S  LAW  163 

Consider  the  proportionality  factor  equal  to  one  and 
write  the  formula. 

Ex.  61.  1000  kilowatts  is  to  be  transmitted  19  miles.  We 
can  transmit  this  at  a  low  pressure  and  large  current  or  by  means 
of  a  small  current  and  at  a  high  pressure. 

(a)  Allow  10  per  cent  of  total  power  for  loss  at  1000  volts. 
Construct  a  table  of  values,  of  current,  voltage,  and  resistance,  for 
transmission  at  1000,  5000,  10,000,  50,000,  and  100,000  volts. 

Conductance.  The  conductance  of  a  wire  is  the  recip- 
rocal of  its  specific  resistance. 

Specific  Resistance.  The  specific  resistance  of  a  material 
is  the  resistance  of  a  unit  cube  of  the  material.  In  the 
C.G.S.  system  the  unit  is  a  centimeter.  '  In  the  English 
system  the  specific  resistance  is  the  resistance  of  a  foot  of 
a  material  having  a  diameter  of  one  mil. 

Ex.  62.  If  the  specific  resistance  of  a  wire  be  taken  as  0.02 
in  practice  instead  of  0.017,  what  error  is  made  if  we  call  the 
conductance  60? 

Ex.  63.  A  sample  of  aluminum  wire  \  in.  in  diameter  under 
test  shows  a  pressure  drop  of  19.2  millivolts  in  a  length  of  3  ft. 
with  a  current  of  100  amps.  What  is  its  specific  resistance  in 
C.G.S.  units  and  what  is  the  resistance  of  a  mil-foot  in  ohms? 

Temperature  Coefficient.  The  temperature  coefficient 
of  a  material  is  the  increase  in  resistance  for  a  temperature 
rise  of  one  degree  centigrade.  The  increase  is  expressed  in 
terms  of  resistance  at  zero  degrees  centigrade: 

R\  is  the  resistance  at  temperature  T\  centigrade; 
R2  is  the  resistance  at  temperature  T2  centigrade; 
Ro  is  the  resistance  at  temperature  zero  centigrade; 
C  is  the  temperature  coefficient; 

(a)  #i=#o(l+CTi), 

(b)  R2  =  Ro(l+CT2). 

Solve  (a)  and  (6)  simultaneously  for  C  by  eliminating  Ro. 


164 


PRACTICAL  MATHEMATICS 


Ex.  64.  The  field  coils  of  a  street  railway  motor  have  a  resist- 
ance of  0.25  ohm  at  a  temperature  of  70°  F.  After  operating 
for  some  time  the  resistance  is  again  measured  and  is  found  to 
be  0.31  ohm.  What  has  been  the  average  rise  of  temperature 
of  the  coils? 

Ex.  65.  The  resistance  of  the  filament  of  an  incandescent 
lamp  at  a  temperature  of  20°  C.  is  50.1  ohms.  When  operating 
under  normal  conditions  at  a  temperature  of  2000°  a  pressure 
of  110  volts  sends  a  current  of  0.37  ampere  through  the  filament. 
What  is  the  average  temperature  coefficient  of  the  filament? 
(This  is  a  tantalum  lamp.) 

Ex.  66.  A  track  is  built  of  90-lb.  rails  (90  lbs.  per  yard). 
The  rails  are  30  ft.  long,  and  the  lengths  are  "bonded"  together 
with  30-in.  lengths  of  round  copper  bonds.  Assuming  the  con- 
ductivity of  steel  to  be  one-twelfth  that  of  copper,  what  should 
be  the  cross-section  of  the  bonds  to  give  a  bond  resistance  equal 
to  1.25  times  a  ft.  of  rail  resistance?  (Steel  rails  weigh  0.280  lb. 
per  cubic  inch.) 

Ex.  67.  If  the  resistance  of  a  mil-foot  of  copper  is  10.8  ohms, 
what  will  be  that  of  a  wire  10  miles  long  and  0.25  in.  diameter? 

Ex.  68.  Variation  of  Resistance  of  an  Incandescent  Lamp 
with  Voltage  Applied  to  it. 

In  Figs.  44  and  45,  L=lamp,  A  =  ammeter,  V  =  voltmeter, 
R  =  variable  resistance,  s  =  switch. 


r-©-JL- 

YJ> 

f*i 

o     o --^- 

< — 

fl 

Fig.  44. 


Fig.  45. 


Derive  the  formulas  for  determining  the  resistance  of  L  in 
both  cases  and  allow  for  the  error  due  to  the  use  of  instru- 
ments. 

Ex.  69.  An  electric  heater  constructed  of  No.  10  iron  wire 
radiates  sufficient  heat  when  the  wire  carries  10  amperes.  The 
heater  is  placed  across  110  volts.  What  is  the  value  of  the  hot 
resistance? 


THE  APPLICATIONS  OF  OHM'S  LAW  165 

Ex.  70.  Temperature  Coefficient. 

r> 

#1  =  ; 77Fi — m)   where   a  =0.004  for  copper,  when  temp,  is 

l+a(l  2-ii 

taken  centigrade  degrees.    An  annealed  copper  conductor  has  a 

resistance  of  15  ohms  at  a  temperature  of    20°  C.    What  will 

the  resistance  be  at  50°  and  also  at  8°  C? 

Ex.   71.     The  observed  resistance  of  a  copper  wire  is  12.74 

ohms  at  85°  F.     What  is  the  resistance  at  65°  F.?     The  factor 

a  =0.0023  for  temperature  F. 

The  unit  of  electric  energy  is  the  work  done  in  one 
second  when  a  current  of  one  ampere  flows  under  a  pressure 
of  one  volt.  This  amount  of  energy  is  called  a  joule  or 
watt-second  (J). 

1  H.P.  hour  =1,980,000  ft.-lbs.; 

1  K.W.  hour  =  3,600,000  watt-seconds  or  joules; 

1  joule  =  .74  ft.-lb. 

Ex.  72.     (a)  J  =PRt.     Solve  for  IR,  t. 

(b)  Substitute  E  for  IR  and  solve  for  E,  I,  t,  J. 

R2 
(d)  Substitute  -j-  for  P  and  solve  for  Q,  t,  R,  J. 

Interpret  each  formula. 

Ex.  73.  What  is  the  amount  of  work  done  in  joules  when  a 
current  of  15  amps,  flows  for  1|  hours  against  a  resistance  of 
2  ohms? 

Ex.  74.  Required,  the  amount  of  work  done  in  joules  in  1 
hour  by  a  current  of  2.5  amperes  under  an  E.M.F.  of  20  volts? 

Ex.  75.  What  is  the  amount  of  work  done  in  45  minutes 
in  a  circuit  having  20  ohms  resistance,  the  E.M.F.  being  110 
volts? 

Ex.  76.     Electric  Power. 

J  =IEt,  joules  =  amp.  Xvolts  Xseconds. 

i  oules 

Divide  both  members  by  time  and  call —  =  watts. 

seconds 

Therefore  watts  =amps.  Xvolts. 

W=IE. 
Interpret  each  formula. 


(c)  Substitute  —  for  P  and  solve  for  /,  E,  R,  t. 


166  PRACTICAL  MATHEMATICS 

Ex.  77.  What  power  in  watts  is  consumed  in  a  circuit  in 
which  0.65  amp.  flows  under  a  pressure  of  110  volts? 

Ex.  78.  In  Ex.  76  substitute  E=IR  and  express  the  value 
of  W. 

Interpret  the  resulting  formula. 

Ex.  79.     Determine  the  power  expended  in  watts  in  an  electric 

circuit  having  a  resistance  of  175  ohms,  through  which  a  current 

of  6  amps,  is  flowing. 

E 
Ex.  80.     In  Ex.  76  substitute  I « —  and  express   the  value 

K 

oiW. 

Interpret  the  resulting  formula. 

Ex.  81.     What  is  the  equivalent  of  1  ft.-lb.  in  terms  of  «/? 

Ex.  82.  Express  in  ft.-lbs.  the  work  accomplished  in  a  circuit 
where  a  current  of  10  amps,  flows  for  1  hour  at  10  volts; 

Ex.  83.  What  is  the  amount  of  work  done  in  ft.-lbs.  by  a 
current  of  5  amps,  flowing  for  20  seconds  against  a  resistance  of 
4  ohms? 

Ex.  84.  What  is  the  equivalent  in  ft.-lbs.  of  1,356,000,000 
ergs.     1  joule  - 10,000,000  ergs. 

British  Thermal  Unit  (B.T.U.)  is  the  quantity  of  heat 
which  will  raise  the  temperature  of  one  pound  of  water  one 
degree  F.  at  or  near  its  temperature  of  maximum  density 
39.1°. 

Ex.  85.  Relation  between  Joules  and  British  Thermal 
Units. 

1  joule  =  .0009477  B.T.U.  Solve  for  one  B.T.U.  and  interpret 
the  resulting  formula. 

Ex.  86.  Given  a  circuit  having  a  resistance  of  2  ohms  through 
which  a  current  of  10  amps,  flows  for  1  hour.  Determine  (a)  the 
work  done  in  joules;  (b)  the  equivalent  in  ft.-lbs.;  (c)  the  number 
B.T.U.  developed. 

Ex.  87.  How  many  B.T.U.  are  developed  in  an  electric  cir- 
cuit having  a  resistance  of  220  ohms  through  which  a  current 
of  0.5  amp.  flows  for  1  mim? 

Ex.  88.  An  electric  circuit  has  a  resistance  of  5  ohms  in  which 
3  amps,  flows  for  12  seconds.  Determine  (a)  the  work  in  joules, 
the  equivalent  ft.-lbs.,  and  B.T.U. 

Pound  Calorie  is  the  quantity  of  heat  that  will  raise  the 
temperature  of  one  pound  of  water  1°  C. 


THE  APPLICATIONS  OF  OHM'S  LAW  167 

A  small  calorie  (H)  is    the  heat    required  to    raise  one 
gram  of  water  from  0°  to  1°  C. 

Ex.  89.     Calorie: 

#  =  .24/22ft  =  ?  joules. 

Solve  for  J  and  interpret. 

Ex.  90.  An  insulated  wire  having  a  resistance  of  10  ohn  s 
is  entirely  immersed  in  water.  How  many  calories  will  be  expended 
in  heating  the  water  when  a  current  of  10  amps,  flows  for  1  hour? 

Ex.  91.     One  calorie  equals  how  many  B.T.U.? 

Ex.  92.    Watt  Hours. 

Express  1  watt-second  in  joules,  calories,  ft. -lbs.,  B.T.U.? 

Ex.  93.  What  is  the  power  in  watts  in  a  closed  circuit  in 
which  electric  energy  is  expended  in  1  hour  to  be  equivalent 
to  33000  ft.-lbs.? 

Ex.  94.  A  power  station  supplies  500  amperes  for  10  hours 
to  a  factory.  The  drop  in  the  line  is  25  volts.  How  much  energy 
in  joules  is  lost  and  what  is  its  equivalent  in  kilowatt  hours? 
What  power  is  used?    What  is  the  efficiency  of  the  transmission? 

Ex.  95.  What  is  the  equivalent  in  ft.-lbs.  spent  in  60  minutes 
for  a  lamp  rated  at  55  watts? 

Ex.  96.  The  electric  energy  expended  in  a  circuit  in  2  hours 
is  equivalent  to  5,000,000  ft.-lbs.  The  E.M.F.  of  the  circuit 
is  110  volts,  what  is  the  current? 

Ex.  97.  An  incandescent  lamp  of  210  ohms  is  to  be  used 
on  a  110-volt  circuit.  Determine  (a)  the  current  required  for 
the  lamp;  (6)  the  watts  consumed;  (c)  how  many  B.T.U.  developed 
per  second;  (d)  how  many  such  lamps  could  be  kept  burning 
by  one  electric  horse  power;  (e)  what  is  the  mechanical  equiva- 
lent of  the  heat  developed  per  second  in  the  lamp;  (/)  for  how 
many  lamps  would  10  KW.  suffice;  (g)  how  many  K.W.  are 
required  to  operate  15  such  lamps? 

Efficiency  of  Transformation  and  Transmission.  Effi- 
ciency is  a  ratio  between  the  useful  work  performed  by  a 
machine  and  the  energy  put  into  it,  in  other  words,  it  is  the 
ratio  of  output  to  input. 

__  .  output 

Efficiency  =  — - — — . 
input 

When  this  ratio  is  multiplied  by  100  the  result  is  ex- 
pressed as  per  cent  efficiency. 


168 


PRACTICAL  MATHEMATICS 


m,       ra  •  r  electrical  output 

The  efficiency  of  a  motor  =  -: — — -. — ; ; — -f^ — — . 

electrical  output + losses 

rpi       rn  ■  £  mechanical  output 

I  he  efficiency  of  a  motor  = : : — : ; . 

mechamcal  output+losses 

The  efficiency  of  a  motor  =  electrical  input -losses 

electrical  input 

Ex.  98.  Write  the  formulas  for  efficiency,  using  the  follow- 
ing notation:   W  =  output,  w  =  losses,  yj  =  efficiency. 

Ex.  99.  The  resistance  of  a  dynamo  armature,  Fig.  46, 
is  0.016  ohm,  that  of  the  external  circuit  0.757  ohm.    The  power 


.757  ft  g 


83.7  Amps. 


Fig.  46. 

required  to  operate  the  machine  is  7.604  H.P.  and  the  current 
produced  is  83.7  amps.     What  is  the  efficiency  of  the  machine? 

Ex.  100.  A  circuit,  Fig.  47,  consists  of  100  inc.  lamps  arranged 
in  20  groups,  each  group  containing  5  lamps  in  series.  The  volts 
between  the  main  wires  at  the  center  of  the  lamp  system  is  550 


10  Groups 


10  Groups 


Res. 
1  Each  Lamp  < 
2121) 


550  V. 


Fig.  47. 

and  the  average  resistance  of  each  lamp  is  21212.  (a)  Deter- 
mine the  watts  used  per  lamp;  (b)  allow  a  line  resistance  of  8 
ohms.    What  is  the  efficiency  of  the  transmission? 

Ex.  101.  What  is  the  commercial  efficiency  of  a  dynamo 
when  a  dynamometer  shows  that  it  absorbs  8  horse-power  of 
mechanical  energy  while  furnishing  92  incandescent  lamps  with 
46  amps,  at  115  volts? 


THE  APPLICATIONS  OF  OHM'S  LAW 


169 


Ex.  102.  The  resistance  of  an  armature  is  0.02  ohm  and  the 
shunt  fields  have  a  resistance  of  25  ohms.  The  generator,  absorbs 
10  horse-power  and  delivers  50  amps,  to  the  line  while  the  brush 
E.M.F.  is  118  volts.     Calculate  the  various  losses.      (See  Fig.  48a.) 

Ex.  103.  Calculate  the  efficiency  of  a  long-distance  line  when 
50  amps,  at  3600  volts  are  supplied  to  it.  The  resistance  of  the 
line  is  10.8  ohms. 


340  K.W, 


Fig.  48. 

Ex.  104.  A  plant,  Fig.  48,  consists  of  a  generator  of  94  per 
cent  efficiency;  step-up  transformers  (lower  to  higher)  96  per 
cent;  line  90  per  cent;  lowering  or  step-down  transformers  95 
per  cent.  The  engine  that  drives  the  dynamo  has  an  efficiency 
of  88  per  cent.  If  340  K.W.  of  energy  is  at  the  secondary  of  the 
step-down  transformer,  calculate  the  plant  efficiency  and  horse- 
power supplied  to  the  engine.    What  is  the  station  efficiency? 


50  Amps. 


Fig.  48a. 


E.M.F.  |  g)   jc.E.M.F. 

Fig.  49. 


Ex.  105.  What  efficiency  will  be  necessary  in  a  stationary 
motor,  Fig.  49,  in  order  that  it  may  develop  450  volts  counter 
E.M.F.  when  the  applied  E.M.F.  is  500  volts? 

Data  : 

Applied  E.M.F.  =500  volts, 

Counter  E.M.F.  =450  volts. 

Fffi  *         _  output  _  counter  E.M.F. 
input  ~  applied  E.M.F. 

Per  cent  efficiency  =  efficiency  X 100. 

Complete  the  problem. 


170  PRACTICAL  MATHEMATICS 

Ex.  106.  50  lamps,  Fig.  50,  are  to  be  supplied  at  110  volts, 
the  hot  resistance  of  each  being  220  ohms.  The  line  loss  is 
5  per  cent;  the  commercial  efficiency  of  the  generator  is  92  per 
cent  and  the  engine  and  belt  losses  15  per  cent.  What  H.P. 
will  be  necessary  to  operate  the  plant? 


Engine  Generator 

Fig.  50. 

Ex.  107.  How  much  power  is  required  to  furnish  40  arc 
lamps,  Fig.  51,  in  series,  with  7  amps,  of  current?  The  resistance 
of  each  lamp  is  8  ohms  and  the  line  25  ohms.  How  many  watts 
are  necessary  per  lamp?  Assuming  a  normal  candle  power  of 
1200  for  each  lamp,  what  is  the  efficiency  in  watts  per  candle? 


8f2  =  res.eachlamp 

7    AmpS.     y  y yf 


Fig.  51. 

Ex.  108.  An  engine  on  full  load  indicates  125  horse-power. 
Assuming  engine  and  belt  losses  to  be  12  per  cent  and  the  com- 
mercial efficiency  of  dynamo  85  per  cent,  line  loss  5  per  cent. 
How  many  110-volt  incandescent  lamps  constitute  the  load? 
What  is  the  voltage  at  the  brushes? 

Net  Work  in  an  Electric  Circuit.    KirchhofFs  First  Law. 

The  sum  of  the  currents  flowing  to  any  point  is  equal  to  the 
sum  of  the  currents  flowing  away  from  that  point. 

Ex.  109.  Construct  a  formula  which  shall  express  the  current 
in'th'e  following  circuit,  Fig.  52,  using  the  following  notation. 


THE  APPLICATIONS  OF  OHM'S  LAW 


171 


E  =  volts  at  the  terminals  of  a  shunt  generator, 
Rf  =  shunt  fielcl  resistance*, 
Ra  =  armature  resistance, 

R  =  resistance  of  lamps  in  the  circuit, 

7=main  current, 

ia  =  armature  current, 

i/ = field  current. 
Determine  /,  ia  and  if,  when  E  =  220  volts,  Rf  =  50tt,  i^«  =  .512? 


www* 


rv^\XK^ 


Fig.  52. 

Kirchhoff's  Second  Law.  In  every  closed  circuit  the 
sum  of  the  products  of  current  and  resistance  taken  round 
the  whole  circuit  equals  the  sum  of  the  E.M.F.  of  the 
circuit.     A  back  or  counter  E.M.F.  is  negative  in  sign. 


Ex.  110. 

E=EM.F.  of  dynamo, 
Eb=  E.M.F.  of  battery, 
/  =  current  through  battery, 


Ra  =  armature  resistance, 
Ri  =  resistance  of  leads, 
Rb=  battery  resistance. 


In  Fig.  53  a  generator,  22  =  116  volts,  is  connected  to  charge 
a  battery  of  50  accumulators  having  an  E.M.F.  of  2  volts  each. 


Fig.  53. 


The  generator  resistance  #a=0.1  ohm,  battery  resistance  Rb=0.18 
and  resistance  of  leads  Rt  =0.12. 

Explain  the  formula  IRa+IRb+IRi  =E-Eb. 

Determine  the  current  I. 


172  PRACTICAL  MATHEMATICS 

Ex.  111.  A  generator,  Fig.  54,  of  135  volts  and"0.01512  charges 
storage  cells  for  seven  hours.  The  resistance  of  leads =0.02512, 
and  53  cells  have  each  0.000212  internal  resistance.  At  starting 
the  E.M.F.  of  a  cell  is  2.1  volts  and  at  the  end  of  the  run  2.35 

.0:5 K 


.0002  ftEach,53  Cells  /"T 

ill  Ml- |||||h-VWWWWWN 


Fig.  54. 


volts.     What  is  the  necessary  regulator  resistance  to  keep  current 
at  200  amps,  at  starting  and  stopping? 

Data: 

#=E.M.F.  generator  =  135, 
Ei  -initial  E.M.F.  battery  =53 X2.1, 
E2  =final  E.M.F.  battery  =53  X2.53, 

R  =  resistance  of  regulator, 
Ra  =.01512 
#6  =53  X. 0002, 
#,  =  .02512. 

E-E1=I(R1+Ra+Rl+Rb). 

135-(2.1x53)=200(fl1  +  .015+.025+[53X.0002]). 

Ri  =  ?  12  regulator  resistance  at  start  of  charge. 

E  —Ei  =I{R<i-\rRa-\-Ri-\-Rb) . 

R2  =  ?  12,  regulator  resistance  at  end  of  charge. 

Ex.  112.  What  is  the  monthly  cost  of  operating  a  10-H.P. 
motor  8  hours  a  day  at  10  cents  per  K.W.  hour? 

Ex.  113.  The  0.11  ohm  leads  fiom  a  50- volt  P.D.  source  are 
carrying  10  amps,  to  an  arc  lamp  of  39  volts,  B.E.M.F.  which 
has  0.09  ohm  resistance  in  the  lamp  coil.  The  carbons  have  resist- 
ances 0.08  and  0.12  ohm  and  the  arc  0.1  ohm.  What  is  the  value 
of  the  adjustable  resistance  which  keeps  the  current  in  the  lamp 
at  10  amps.?    See  Figs.  55  and  56. 


THE  APPLICATIONS  OF  OHMS  LAW 


173 


Fig.  55  shows  an  arc  lamp  in  which  s=  switch,  OK  =  clutch, 
DP  =  dash  pot,  T=  trip,  M=  solenoid  magnet,  CC  =  carbons, 
AR=  ad  just  able  resistance. 

Ex.  114.  When  a  storage  battery,  Fig.  57,  a  heavy  rheostat  to 
prevent  short  circuiting,  an  ammeter,  and  an  armature  whose  resist- 
ance was  sought,  were  joined  in  series,  the  current  was  15  amps. 
The  voltmeter  bridged  across  the  two  commutator  bars  in  contact 
with  the  brushes  showed  0.09  volt.    What  is  the  armature  resistance? 


Fig.  55. — Arc  Lamp. 


r— I- 


X.09  8 


Fig.  56. 


rOr^ 


Amps. 


00  V 

L-®-l 
Fig.  57. 


A  voltmeter  may  be  used  to  test  a  joint,  switch  contact, 
battery  connection,  etc.     The  working  formula  is 


R  = 


V 
V 


where  I  is  the  strength  of  current  passing  through  the  con- 
nection or  joint  and  V  the  voltage  drop  across  it,  and  R  its 
corresponding  resistance. 


174 


PRACTICAL  MATHEMATICS 


Ex.  115.  A  foot  of  search-light  wire  has  a  poor  connection. 
It  shows  a  drop  of  .1  volt,  whereas  a  foot  of  regular  main  shows 
^V  v°lt  while  carrying  100  amperes.  What  is  the  resistance  of 
the  joint  and  the  power  lost  in  it? 

Insulation  resistance  may  be  measured  by  connecting 
it  to  a  well-insulated  generator  or  battery  in  series  with  a 
high  resistance  Weston  voltmeter.  These  are  designated 
by  R,  B,  and  r  respectively,  and  the  corresponding  volt- 
meter reading  is  v.  The  voltmeter  is  then  shunted  by  the 
switch  S,  as  shown  in  the  figure  (58).     Upon  closing  the 


^ 


Fig.  58. 

switch  the  voltmeter   reading    becomes  V.     The   working 
formula  is 


R 


V—vr 
v 


Ex.  116.  The  deflection  on  a  20000-ohm  voltmeter  indicated 
100  volts  when  connected  to  a  generator.  An  unknown  resistance 
was  inserted  in  the  circuit  and  the  voltmeter  then  indicated  40 
volts.    What  was  the  value  of  the  unknown  resistance? 

Ex.  117.  The  deflection  of  a  voltmeter  was  120  when  the  150 
volt  coil  was  connected  to  a  well-insulated  storage  battery.  The 
3  volt  coil  of  the  instrument  was  then  connected  between  a  resist- 
ance R  and  the  battery  and  the  corresponding  deflection  indicated 
■fo  volt.    What  was  the  resistance  of  R? 

Ex.  118.  What  current  passes  through  a  1.5-ohm  electro- 
magnet which  is  connected  to  a  cell  having  2  volts  E.M.F.  and 
.5  ohm  resistance? 

Ex.  119.  A  cell  of  2  volts  and  internal  resistance  0.5  ohm 
is  joined  in  series  with  a  number  of  different  resistance  spools. 
The  drop  over  a  0.4  ohm  spool  is  0.6  volt.    What  is  the  current 


THE  APPLICATIONS  OF  OHM'S  LAW  175 

flowing  through  the   circuit?    What  part  of  the  total  E.M.F. 
is  used  in  overcoming  the  resistance  of  the  cell? 

Ex.  120.  A  cell  with  an  internal  resistance  of  2  ohms  sends 
a  current  of  0.035  amp.  through  an  electromagnet  of  a  bell  having 
a  resistance  of  48  ohms.    What  is  the  E.M.F.  of  the  cell? 

The  diameter  of  copper  wire  required  for  feeders,  mains, 
branches,  service  wires  or  inside  wiring  depends  upon  the 
following: 

n  =  number  of  lamps  in  parallel ; 
7=the  current  required  for  each  lamp; 
Z  =  the  distance  of  the  lamps  from  the  center  of  distribu- 
tion expressed  in  feet; 
j£  =  the  total  drop  in  the  wires; 
d  =  the  diameter  of  the  wire; 


-4 


21.6n/Z 


E 


Ex.  121.  Compute  the  diameter  and  determine  the  size  of 
copper  wire  which  is  required  to  supply  50  16  c.p.  lamps  at  a 
distance  of  150  ft.  Each  lamp  is  to  burn  at  110  volts  and  con- 
sume 53  watts.    The  drop  in  the  leads  is  2  volts. 

Ex.  122.    Watts  per  candle. 


1  = 


Wicp  Wi  =  number  of  watts  per  candle, 

E   '  cp  =  candle  power  of  lamps. 


Solve  for  cp  and  interpret  the  formula. 

Solve  for  Wi  and  interpret  the  formula. 

Ex.  123.  Use  the  preceding  formula  (Ex.  122)  to  determine 
Wi  when  I  =2.15,  E  =220,  and  cp  =32. 

Ex.  124.  Use  the  formula  in  Ex.  122  to  determine  Wi  when 
7  =  .5,  #  =  110,  andcp  =  16. 

The  number  of  volts  drop  in  lamp  wires  is  given  by  the 
formula, 

xE 


v  = 


100 -X' 


176  PRACTICAL  MATHEMATICS 

where  *>  =  the  number  of  volts  drop  in  the  wires; 
E  =  voltage  delivered  to  a  lamp; 

z  =  a  whole  number  and  is  the  percentage  drop  or  100 
times  the  ratio  of  the  drop  to  the  voltage  received 
by  the  lamps- 
Ex.  125.     The  leads  to  a  cluster  of  110- volt  lamps  are  figured 
for  a  5  per  cent  drop.    What  is  the  actual  volts  lost  in  the  leads? 

Ex.  126.  What  size  of  wire  will  carry  50  amps.  100  ft.  to  a 
110-volt  motor  with  a  drop  of  2  per  cent? 

Ex.  127.  In  the  preceding  formula  solve  for  x,  and  express 
the  per  cent  drop  in  wires  in  terms  of  volts  drop  and  volts 
delivered. 

Ex.  128.  There  is  5  volts  drop  in  the  leads  connected  to  a 
motor  running  at  an  E.M.F.  of  105  volts.  What  is  the  per  cent 
drop? 

Ex.  129.  A  group  of  110-volt  lamps  is  to  be  fed  by  means 
of  a  cable  whose  length  is  50  yds.  The  voltage  drop  is  not  to 
exceed  about  2  volts.  Allow  55  watts  per  lamp.  What  size  wire 
should  be  used? 

Ex.  130.  A  current  of  35  amps,  is  conducted  150  yds.  The 
maximum  drop  allowed  is  3  per  cent.  Calculate  the  cable  so 
that  220  volts  are  delivered  to  the  receiving  circuit? 

Ex.  131.  A  current  of  20  amps,  is  to  be  conducted  150  yds. 
The  voltage  drop  allowed  is  6  volts.  Determine  the  size  of  the 
cable. 

Ex.  132.  A  stationary  armature,  resistance  0.03  ohm,  was 
accidentally  connected  with  a  110-volt  circuit.  Estimate  the 
amount  of  current  causing  the  illumination. 

A  storage  battery,  motor,  or  arc  lamp  exerts  a  back 
pressure  when  supplied  by  a  generator.  If  the  generator 
pressure  =  E  and  the  back  pressure  =  e,  then  the  effective 
pressure  =  E—e  and  therefore  the  current  flowing  through 
the  circuit  is  given  by  the  formula, 

7    E-e 
1       R  ' 

Ex.  133.  Consider  a  simple  circuit  containing  a  generator  of 
3  volts  and  0.02  ohm,  a  storage  cell  of  2  volts  and  0.005  ohm, 


THE  APPLICATIONS  OF  OHM'S  LAW  177 

and  leads  of  0.1  ohm.  What  current  flows  through  this  circuit 
and  what  is  the  drop  at  the  different  points? 

Ex.  134.  What  is  the  E.M.F.  of  a  generator  of  0.02  ohm 
supplying  100  amps,  to  54  cells  in  series,  each  of  which  has  2.3 
volts  back  E.M.F.  and  0.004  ohm  resistance?  The  leads  resistance 
=  0.03  ohm.  What  is  the  voltage  at  the  generator  and  at  the 
battery  terminals? 

Ex.  135.  A  10-mile  arc-light  circuit  of  No.  6  B.  &  S.  wire  has 
10  amperes  flowing  through  it.  What  horse-power  is  lost  in  the 
circuit? 

The  diameter  d  of  a  wire  required  to  transmit  HP 
horse-power  over  a  distance  of  L  feet  with  v  volts  loss  in  the 
wire  to  a  motor  requiring  E  volts  and  having  an  efficiency 
of  iq  is  given  by 


-4 


746  HP  2  L  10.8 
vE-q 


Simplify  the  numeric  part  of  this  expression  and  show 
how  it  will  have  to  be  modified  in  order  to  apply  to  a  50 
per  cent  overload  on  the  motor. 

Ex.  136.  A  110-volt  hoist  motor  of  15  horse-power  is  75  ft. 
from  the  main  switch.  Select  the  tap  wires  to  allow  a  drop  of 
3  volts  from  the  switch  to  a  motor  of  90  per  cent  efficiency. 

A  constant  current  motor  of  tq  efficiency  requires  E  volts 
and  /  amperes  to  give  HP  horse-power  as  follows: 

EjmHP 
Iri 

Solve  for  tq  and  interpret  the  resulting  formula. 

Ex.  137.  What  current  will  be  delivered  to  supply  a  220-volt 
motor  of  90  per  cent  efficiency  in  order  to  give  15  H.P.? 

Ex.  138.  Make  a  sketch  of  the  wiring  and  the  distribution 
of  the  current  in  all  the  buildings  of  the  Institute.  Determine 
the  drop  at  different  loads  by  communicating  with  the  engineer 
at  the  generating  plant.  Work  out  a  complete  wiring  chart  for 
the  system. 

Ex.  139.  In  a  transmission  line  three  hard-drawn  copper 
wires  are  used.    They  have  97  per  cent  of  the  conductivity  of 


178  PRACTICAL  MATHEMATICS 

pure  copper.  The  wire  is  0.182  in.  in  diameter  and  there  are 
99.8  ft.  of  it  in  a  pound.  Determine  the  following  items:  (a) 
tensile  strength  of  copper  line;  (b)  tensile  strength  of  an  aluminum 
line  of  the  same  conductivity;  (c)  feet  per  pound  of  this  alumi- 
num wire;  (d)  diameter  of  the  aluminum  wire;  (e)  price  per 
pound  of  an  aluminum  wire  which  shall  be  equivalent  in  con- 
ductivity to  copper. 

Ex.  140.  A  wattmeter  fluctuates  between  7000  and  5000 
for  5  minutes.    What  is  the  value  of  the  average  watt  hour? 

Ex.  141.  A  bus  of  copper  3|Xf  in.  is  35  ft.  in  length,  (a) 
What  is  its  resistance  at  20°  C?  K  =  10A.  (b)  What  length 
of  bus -will  serve  as  a  shunt  for  a  2000  amp.  ammeter  requiring 
50  millivolts  for  full-scale  deflection. 

Ex.  142.  A  90-lb.  steel  rail  is  30  ft.  in  length.  K  for  steel 
ia  8  times  that  of  copper  and  1  cu.in.  weighs  0.28  lb.  (a)  What 
is  the  resistance  of  the  rail?  (b)  What  is  the  resistance  of  a 
mile  of  single  rail  allowing  5,per  cent  additional  for  the  bonds? 

Ex.  143.  A  copper  transmission  line  has  a  resistance  of  4.5 
ohms  at  70°  F.  What  is  the  range  of  resistance  for  temperature 
varying  between  -25°  F.  to  110°  F.? 

Ex.  144.  A  copper  field  coil  has  a  resistance  =25  ohms  at 
70°  F.  After  running  the  machine  four  hours  the  resistance  of  the 
coil  increased  4  ohms.  What  was  the  corresponding  temperature 
of  the  coil? 


CHAPTER  VIII 

EFFICIENCY  OF  GENERATORS  AND  MOTORS 

1.  Efficiency  is  the  ratio  of  the  output  to  the  input 
of  any  machine  or  device. 

output 


Efficiency 


input 


The  electrical  efficiency  (yj«)  of  an  electric  machine  is 
the  ratio  of  the  useful  energy  to  the  total  electric  energy 
(W)  in  its  armature.  The  electric  energy  in  the  armature 
coils  consists  of  the  useful  energy  (Wu)  plus  the  energy 
(Wa)  loss  due  to  armature  resistance  and  the  energy  (W/) 
loss  due  to  the  field  resistance. 

The  electrical  efficiency  of  a  generator  is  given  in  (1). 

,  -x  _^"_  ^ output 

^'W'Wu+Wa+Wf  "output -{-copper  losses* 

The  electrical  efficiency  of  a  motor  is  given  in  (2), 

f<y\  -  ^u  —  W—(Wa-\-Wf)  _  input -copper  losses 

{Z)         r]e~W~~~     ~W  input  /  ' 

In  the  case  of  a  generator  WU  =  EI,  and  in  the  case  of 
a  motor  W  =  EI,  in  which  E  is  the  brush  voltage. 

The  electrical  efficiency  does  not  include  waste  due  to 
hysteresis,  or  eddy  current  losses,  nor  friction,  but  is  depend- 
ent upon  the  copper  losses  only. 

Ex.  1.  Express  the  W's  in  terms  of  E,  I,  and  R  and  substitute 
in  (1)  and  (2).  Write  the  formulas  for  the  electrical  efficiency 
of  (a)   a  series-wound  generator;    (6)   a  shunt-wound  generator; 

179 


180  PEACTICAL  MATHEMATICS 

(c)  a  compound-wound  generator;    (d)  a  series-wound  motor;    (e) 
a  shunt-wound  motor;   (/)  a  compound- wound  motor. 

Owing  to  the  ambiguity  of  the  word  efficiency  when  not 
specifically  defined  there  is  a  widely  accepted  designation 
known  as  the  net  efficiency. 

2.  The  commercial  or  net  efficiency  of  an  electric 
machine  is  the  ratio  of  the  output  to  the  intake.  The 
intake  (Wi)  of  a  generator  equals  the  input  (W)  or  total 
energy  generated  in  the  armature,  plus  the  energy  (Wn) 
losses  due  to  hysteresis,  eddy  currents  (We)  and  to  friction 
(F).  The  intake  of  a  motor  is  the  electric  energy  delivered 
at  its  terminals.  The  output  of  a  generator  is  the  available 
electrical  energy  at  its  terminals.-  The  output  of  a  motor 
is  the  available  mechanical  energy  at  its  shaft.  The  com- 
mercial efficiency  (tqc)  is  given  in  (3) 


d\        —Ejf  — ^ output 

W     ^-Wj-W+Wh+We+F  "intake 


Wu 


Wu+W*+Wf+Wn+We+F' 

The  commercial  efficiency  of  a  motor  is  given  in  (4), 

Wu       Wj-iWa  +  Wf+Wn  +  We  +  F) 


(4)       T)C  = 


Wi  Wj 

_  intake  —  all  losses 

intake 


Ex.  2.  Express  the  copper  losses  in  terms  of  E,  I,  and  R  and 
substitute  in  (3)  and  (4).  Write  the  formulas  for  the  commer- 
cial efficiency  of  (a)  a  series-wound  generator;  (6)  a  shunt- wound 
generator;  (c)  a  compound-wound  generator;  (d)  a  series- wound 
motor;    (e)  a  shunt-wound  motor;    (/)  a  compound-wound  motor. 

3.  The  gross  efficiency  (%)  of  an  electric  machine 
is  the  ratio  of  the  commercial  efficiency  to  the  electrical 
efficiency. 

(5)  r»=\ 


EFFICIENCY  OF  GENERATORS  AND  MOTORS     181 

Ex.  3.  Express  t\g  for  a  generator  and  for  a  motor  both  in 
terms  of  W  and  also  in  terms  of  E,  I,  and  R. 

Ex.  4.  Determine  the  electrical  and  commercial  efficiencies 
of  a  150- volt  8000  ampere  multipolar  shunt  generator.  The 
armature  resistance  equals  105  microhms,  and  the  total  exciting 
current  for  the  field  magnets  equals  13.9  amperes.  The  hysteresis 
loss,  eddy  current  loss,  and  friction  loss  are  11.22  K.W.,  .58  K.W., 
and  40  K.W.  respectively. 

Ex.  5.  Determine  the  electrical  and  commercial  efficiencies 
of  a  540-volt  3700-ampere  multipolar  compound  generator.  The 
armature  resistance  equals  .00216  ohm;  the  total  exciting  current 
for  the  shunt  fields  equals  20.8  amperes;  the  resistance  of  the 
series  field  equals  147  microhms.  All  other  losses  aggregate 
69.7  kilowatts. 

Ex.  6.  Determine  the  electrical  and  commercial  efficiencies 
of  a  bipolar  low-speed  compound  50  K.W.  125-volt  generator.  The 
armature  resistance  equals  .0314  ohm;  the  total  current  required 
for  the  shunt  field  equals  2.84  amperes;  the  resistance  of  the  series 
field  equals  .00106  ohm.  The  hysteresis  and  eddy  current  losses 
total  332  watts  and  the  friction  losses  2.5  kilowatts. 


CHAPTER  IX 
THE  ALGEBRA  OF  THE  MAGNETIC  CIRCUIT 

1.  Comparison  of  Magnetic  and  Electric  Circuit  Rela- 
tions. In  the  magnetic  circuit  the  elements  are  the  mag- 
netomotive force,  M.M.F.(M);  the  reluctance  (R);  and  the 
flux  (<£).  These  elements  enter  into  a  relation  which  is 
analogous  to  the  relations  of  the  elements  E,  R,  and  I  of 
Ohm's  Law. 

Magnetic  Electric 

(1)    -%  (2)    /-| 

(o\     -pi      _  magnetomotive  force  _  magnetic  pressure 
reluctance  magnetic  resistance* 

Transform  (1)  and  solve  for  M  and  R  and  interpret  (1) 
and  the  resulting  equations. 

2.  The  intensity  of  a  magnetic  field  or  magnetizing 
force  (H)  is  compared  with  the  PD  or  drop  in  an  electric 
circuit,  where  I  is  the  length  of  either  the  magnetic  or  electric 
circuit.    H  is  called  the  magnetic  drop. 

Magnetic  Electric 

(4)     H=^-.  (5)     PZ>=|. 

Transform  (4)  solving  for  M  and  I  and  interpret  (4) 
and  the  resulting  equations. 

3.  The  inductance  or  flux  density  (B)  is  the  number  of 
lines  of  force  per  unit  area  and  is  compared  to  the  current 

182 


THE_  ALGEBRA  OF  THE  MAGNETIC  CIRCUIT    183 

density  (Id)  per  unit   area  where  A  is  the  area  of  either 
the  magnetic  or  electric  cross-section. 

Magnetic  -       Electric 

(6)     B=|.       .  (7)     U=-A. 

Transform  (6)  solving  for  B  and  A  and  interpret  (6) 
and  the  resulting  equations.  If  A  is  expressed  in  square 
centimeters,  then  B  is  the  density  per  square  centimeter, 
but  if  A  is  expressed  in  square  inches,  then  B  is  the  density 
per  square  inch. 

4.  The  reluctance  (R)  of  a  magnetic  circuit  corresponds 
to  the  resistance  of  an  electric  circuit  and  like  resistance 
is  dependent  upon  the  length,  cross-section  and  specific  nature 
of  the  material. 

Magnetic  Electric 

(8)     £=-/.  (9)     R=f. 

(10)     R  =  -^=-d  (11)     £*-L--'4- 

v     J  [lA     ^  A  gA      p  A 

K  is  called  the  specific  resistance  or  resistivity ; 
K\  is  called  the  specific  reluctance  or  reluctivity ; 
p  is  called  the  specific  conductance  or  conductivity; 
pi  is  called  the  specific  permeance  or  permeability. 

Transform  (8)  and  (10)  and  interpret  (8)  and  (10)  and 
the  resulting  equations. 

5.  In  a  magnetic  circuit  the  reciprocal  of  reluctance  is 
called  permeance  (P)  and  corresponds  in  the  electric  circuit 
to  conductance  (G)  which  is  the  reciprocal  of  resistance : 

Magnetic  Electric 

(12)    P=|.  (13)    G=l 


184  PKACTICAL  MATHEMATICS 

Substitute  the  value  of  R  from  (10)  in  (1)  and  then 
replace  ■—  by  B  from  (6).     Interpret  the  resulting  equation. 

A. 

6.  The  Field  within  a  Coil.  A  solenoid  is  a  helix  or 
coil  of  wire  which  creates  a  magnetic  field  when  a  current 
of  electricity  passes  through  it.  The  strength  of  the  mag- 
netic field  (H)  within  its  interior  is  directly  proportional 
to  the  total  number  (N)  of  turns  of  wire  and  to  the  current 
passing  through  it  and  is  inversely  proportional  to  its 
length  measured  in  centimeters. 

„    NI  H       Nlh  „     Hxh  NI 

I  Hi     Nihl  Nih   I 

4x  Hih 

7.  The    proportionality    factor    equals   —  =  1.26  =  ——- 

10  Nili 

which  is  the  field  intensity  in  the  air  within  the  solenoid 
when  it  is  wound  with  1  turn  per  centimeter  of  length  and 
has  1  ampere  of  current  flowing  through  it.     Therefore, 

(14)  H=^=^l=h2Qf. 

The  product  NI  is  termed  the  ampere-turns.    Interpret 

NI 

(14)  in  terms  of  ampere-turns.     The  quantity  —  is  called 

o 

the  ampere-turns  per  centimeter  of  length.  (14)  may 
be  rewritten  (15),  in  which  the  product  HI  is  called  the 
magnetomotive  force  and  is  abbreviated  by  (M)  according 
to  (4). 

(15)  Hl=1.2QNI  =  M. 

8.  Iron  or  any  other  magnetic  material,  when  inserted 
within  the  solenoid,  has  the  effect  of  increasing  the  number 
of  lines  of  force  from  (H)  per  square  centimeter  to  a  flux 
density  of  B  lines  per  square  centimeter,  i.e.,  B  gausses.    The 


THE  ALGEBRA  OF  THE  MAGNETIC  CIRCUIT     185 

ratio  of  B  to  H  is  called  the  permeability  of  the  magnetic 
material  and  is  designated  by  (jx). 

(16)  H  =  §, 

solve  (16)  for  H  and  for  B. 

H  is  called  the  magnetizing  force  and  is  the  field  intensity 
in  air,  i.e.,  the  number  of  lines  of  force  per  square  centimeter 
in  air.  What  is  the  distinction  between  magnetizing  force 
(H)  and  magnetomotive  force  (M).  The  permeability  of 
a  magnetic  substance  is  a  variable  quantity,  i.e.,  for  each 
definite  value  of  B  there-  is  a  definite  value  of  [l.  These 
values  are  determined  in  Ex.  10. 

There  is  no  electric  analogue  for  equation  (16). 

9.  The  reluctance  (R)  of  a  compound  circuit  such  as 
the  magnetic  circuit  of  a  generator  may  be  obtained  by  adding 
the  individual  reluctances  in  the  path  of  the  flux.  Ri,  R2, 
R3  correspond  respectively  to  the  reluctances  of  the  magnet 
frame,  air-gap  and  core  of  the  armature.  Reluctances  in 
series  (17a)  are  united  like  series  resistances  and  reluctances 
in  parallel  (176)  are  united  like  resistances  in  parallel. 

(17a)  R  =  Ri+R2+Rs, 

(176)  i=zr+7r+7r> 

K     it/i     til     txz 

M  ^^ 

show  that  <£  =  p    ,  p    ,  p    and  interpret. 
Ki-\-£i2  -1-/13 

10.  The  total  ampere-turns  (NI)  required  for  the 
magnetic  circuiMs  the  sum  of  the  Nh,  NI2,  NI3,  which 
represent  the  ampere-turns  required  to  overcome  the  reluc- 
tance of  the  magnet  frame,  the  armature  core  and  the  air 
gap  spaces  respectively,  NI4  represents  the  ampere-turns 
required  to  compensate  for  the  armature  reaction. 

(18)  NI=NT1+NT2+Nh-\-Nh. 


186  PBACTICAL  MATHEMATICS 

Ex.  1.  Show  that  (14)  becomes  (19)  when  H'  is  given  in  lines 
per  square  inch  and  V  in  inches. 

(19)  NI  =  -r-^^H'l'  =  .3133  JF1\ 

4x2.54 

One  maxwell  means  one  line  of  force;  one  gauss  equals  one 
maxwell  or  one  line  of  force  per  square  centimeter. 

Ex.  2.  (a)  What  is  the  meaning  2.5  kilomaxwells ;  (6)  What 
is  the  meaning  of  3.5  megamaxwells;  (c)  What  is  the  meaning  of 
2500  gausses;  (d)  How  many  gausses  are  there  in  2500  maxwells 
per  square  inch? 

Ex.  3.  In  a  magnetic  circuit  the  cross-section  is  f  in.  X?  in.,  and 
the  magnetic  density  is  50000  lines  of  force  per  square  inch. 
How  many  lines  of  force  thread  the  circuit? 

The  area  =*f  Xl  -  .375  sq.in. 

3>  =  BA  =50000 X. 375  =  18750  lines  thread  the  circuit. 


Ex.  4.  The  cross-section  of  a  magnetic  circuit  is  circular  and 
1.5  cm.  in  diameter.  The  magnetic  density  is  3000  lines  per  square 
centimeter.  What  is  the  total  number  of  lines  threading  the 
circuit? 

Ex.  5.  How  many  maxwells  per  square  centimeter  are  there 
in  a  round  bar  magnet  i  in.  in  diameter  when  4500  lines  of  force 
pass  through  it? 

Ex.  6.  How  many  gausses  are  there  in  a  bar  magnet  2  cm. 
X.75  cm.  when  9000  lines  of  force  pass  through  it? 

Ex.  7.  The  magnetic  density  in  a  bar  magnet  \  in.  X|  in.  is 
40000  lines  of  force  per  square  inch.  What  total  number  of  lines 
thread  the  magnet? 

Ex.  8.  The  permeability  of  a  piece  of  iron  is  850,  when  the 
magnet  density  is  59500  lines  of  force  per  square  inch.  What  is 
the  field  density  required  to  produce  that  magnetic  density? 

Ex.  9.  The  magnetizing  force  acting  on  a  piece  of  iron  is  600, 
and  the  magnetic  density  produced  is  54300  lines  of  force  per  square 
inch.    What  is  the  permeability  at  that  stage  of  magnetization? 

Ex.  10.  Prepare  a  table  of  values  between  B,  H,  and  (x,  using 
(16)  in  connection  with  (20),  (21),  (22),  (23),  and  (24).  Assume 
values  of  B  beginning  at  2000  and  extending  to  25000  in  incre- 
ments (increases)  of  500  in  each  step. 


THE  ALGEBRA  OF  THE  MAGNETIC  CIRCUIT    187 

Permeability  for  low  combined  cast  iron: 
(20)  .     ,=  380-2»^. 

Permeability  for  high  combined  carbon  cast  iron: 

(2!)  ,-  200-6-»^. 

Permeability  for  malleable  cast  iron: 

15(6000  -B)* 


(22)  ti=  700 


106 

Permeability  for  cast  steel: 

(23)  ,  =  1200-»^. 

Permeability  for  wrought  iron: 

3.2(7500-5)2 


(24)  ^=2800 


103 


Ex.  11.  A  bar  magnet  of  low  combined  cast  iron  has  a  cross- 
section  of  l|  ins.  and  has  a  mean  length  of  18.25  ins.  How  many 
ampere-turns  are  required  to  force  7000  lines  through  the  magnet 
if  the  air-gap  between  the  magnet  and  its  armature  is  negligible? 
How  will  the  result  alter  if  the  air-gap  is  f  in.  in  length?  Use 
formula  (19)  in  Ex.  1  for  English  units. 

The  pull  (F)  in  pounds  of  an  electromagnet  is  given 
in  (25)  where  B  is  the  magnetic  density  per  square  inch  and 
A  the  polar  area  in  square  inches. 

(25)  F  =  t 


72134000" 


Ex.  12.    What  is  the  pull  on  the  magnet  in  Ex.  10  when  the 
current  is  .25  ampere? 


188  PRACTICAL  MATHEMATICS 

11.  In  calculating  magnetic  leakage,  the  total  number 
of  lines  of  force  (<£)  is  expressed  in  terms  of  the  useful  lines 
(<!>«),  the  stray  lines  (<£s),  the  percentage  of  leakage  by 
(p)  and  the  coefficient  or  allowance  for  leakage  by  (X). 

(26)  $=$S+<IV 

100$s 


(27)  p  = 

(28)  X  = 


3> 


Transform  (26),  (27),  (28)  and  interpret  the  resulting 
equations. 

Ex.  13.  Assuming  that  the  magnetic  leakage  in  an  electro- 
magnet is  25  per  cent,  and  that  there  are  75000  useful  lines  of 
force,  how  many  lines  of  force  are  provided  by  the  magnetizing 
coil? 

Ex.  14.  100000  lines  of  force  are  produced  by  the  magnetizing 
coils  of  an  electromagnet  of  which  84000  are  useful.  What  is 
the  percentage  leakage?    What  is  the  coefficient  of  allowance? 

Ex.  15.  In  an  electromagnet  there  are  27000  stray  lines  of 
force  and  63000  useful  lines.    What  is  the  percentage  of  leakage? 

Ex.  16.  The  magnetic  leakage  in  an  electromagnet  is  15  per 
cent  and  there  are  110000  useful  lines  of  force.  How  many  lines 
are  produced  in  the  magnetizing  coils? 

Ex.  17.  The  magnetic  leakage  in  an  electromagnet  is  35  per 
cent.  There  are  60000  lines  produced  by  the  magnetizing  coils. 
How  many  lines  are  useful? 

12.  Leakage  Permeance  between  two  surfaces.  From 
(10)  and  (12)  we  have  the  law  for  the  permeance  of  an  air- 
gap  expressed  in  C.G.S.  units  in  (29) .  Since  the  permeability 
of  air  equals  1  we  may  write: 

1  _       mean  area  of  exposed  surfaces 
(29)  R~ mean  length  of  path  between  surfaces 


THE  ALGEBRA  OF  THE  MAGNETIC  CIRCUIT     189 


In  the  case  of  two  parallel  surfaces,  Fig.  59,  of  approx- 
imate equal  area  the  permeance  is  given  in  (30)  for  C.G.S. 
units, 


(30) 


P  = 


2d 


Write  the  formula  for  the  permeance  when  English 
units  are  to  be  substituted. 

When  the  two  rectangular  surfaces  lie  in  the  same  place 
with  their  edges  parallel  as  shown  in  Fig.  60  the  leakage 
paths  will  be  semicircular  for  a  small  separation  of  the 


Fig.  60. 


edges  and  the  permeance  is  given  in  (31)   and   (31a)  for 
C.G.S.  units,  (31)  is  an  approximation, 


(31)     P  = 


2di       x 
d2-di2 


(31a)     P  =  ±\o&%. 
x  a\ 


Ex.  18.  Determine  the  value  of  P  when  a  =  8  ins.,  d2=3  ins., 
dx  =  1  in. 

Substitute  in  (31),  (31a)  and  (316).  When  the  separation 
between  the  edges  is  considered,  the  path  of  the  flux  is  made 
of  arcs  joined  by  straight  lines   and  (31a)  is  modified  into 

(316), 


(316) 


a  ,       If  xa*2  ,  0 


190 


PRACTICAL  MATHEMATICS 


When  the  two  surfaces  are  at  right  angles  as  shown  in 
Fig.  61,  which  represents  the  rotation  of  the  left-hand  plane 
upward  with  a  radius  dts  then  the  permeance  formula  is 
given  in  (32)  and  (32a),  (32)  is  an  approximation, 

w  pJJ!0&-    (32a)  ^%--!{f+2-+ 

13.  Fig.  62  illustrates  the  path  of  the  flux  in  a  gen- 
erator where  3>  is  the  total  flux  in  the  air-gap.     The  mean 


Fig.  61. 


Fig.  62. — Generator  Flux. 


length  of  the  magnetic  circuit  in  the  frame,  poles,  air-gaps, 
and  core  are  designated  by  lh  2lp,  2la  and  lc  respectively. 
The  corresponding  cross-sections  are  Af,  Ap,  Aa  and  Ac  and 
the  respective  flux  densities  are  Bf,  Bp,  Ba  and  Bc. 

Assuming  no  leakage  interpret  (33)  and  (34)  INf,  IN 
INa,  INc  are  the  respective  ampere-turns  per  unit  length. 


pj 


(33) 


* = AtBf =ApBp  =  AaB„  =  AcBc 


(34) 


IN=INJ.f+2INjp,+2INala+INclc. 


THE  ALGEBEA  OF  THE  MAGNETIC  CIRCUIT     191 

14.  The  Energy  Loss  Due  to  Hysteresis.  The  retention 
or  lagging  of  part  of  the  magnetic  field  in  the  iron  of  arma- 
ture cores  and  alternating  current  devices  is  called  hysteresis. 
A  definite  expenditure  of  energy  (Ph)  in  ergs,  is  required 
to  reverse  or  reestablish  a  varying  flux  through  a  magnetic 
circuit.  Ph  is  directly  proportional  to  the  nth  power  of  the 
maximum  value  of  the  flux  density  (Bmax) ,  also  to  the  volume 
(V)  of  magnetic  material  in  cubic  centimeters,  and  to  the 
number  of  cyclic  changes (/)  or  complete  reversals  per  second. 

Pn*VfBnmax, 

(35)  Ph  =  KVfBnmax  ergs. 

Dr.  Steinmetz  and  his  associates  use  1.6  for  the  average 
value  of  the  exponent  n.  The  proportionality  factor  K 
is  called  the  hysteretic  resistance  and  its  average  value 
for  sheet  iron  is  .0035. 

Ex.  20.  Show  that  (35)  becomes  (36)  when  the  Ph  expresses 
the  energy  lost  in  ergs  for  1  cu.cm.  per  1  cycle  per  second. 

(36)  Ph=KB16maxer^. 

Show  that  (35)  reduces  to  (37)  when  Ph  expresses  the  energy 
in  watts,  V  the  mass  in  cubic  feet,  Bmax  the  flux  density  per  square 
inch,  assuming  K  -  .0035. 

(37)  P,=5.19xlO-7Fj5L6max. 

Calculate  the  values  of  Ph  from  (36),  assuming  i£  =  .0035,  and 
the  values  of  .Bmax  ranging  from  1000  to  12000  lines  per  square 
centimeter  in  gradations  of  1000.  Use  a  slide  rule  as  a  check  and 
tabulate  the  values  for  Bmax,  Bh6max,  and  Ph. 

Ex.  21.  Calculate  the  values  of  Ph,  from  (37),  assuming 
/  =  1,  and  7  =  1,  and  values  of  Bmax  ranging  from  10000  to  100000 
lines  per  square  inch,  in  gradations  of  10000. 

Ex.  22.  What  is  the  hysteresis  energy  loss  in  a  machine 
containing  3000  cu.cm.  of  iron?  The  number  of  cycles  equals 
60,  K  =  .0035,  and  Bmax  =  10000. 

Ex.  23.  The  magnetic  material  in  a  machine  has  a  cross- 
section  of  10  sq.in.  and  a  mean  length  of  30  ins.    There  are  45000 


192  PRACTICAL  MATHEMATICS 

lines  per  sq.in.  and  K  =  .0035.    What  is  the  value  of  the  hysteresis 
loss  for  25  and  also  for  60  cycles? 

15.  The  Energy  Loss  Due  to  Eddy  Currents.  The 
motion  of  the  core  of  an  armature  through  a  magnetic  field 
induces  local  or  eddy  currents  within  it.  The  energy  loss 
(Pe)  in  ergs,  due  to  eddy  currents  is  directly  proportional 
to  the  square  of  the  maximum  value .  of  the  flux  density 
(Bmax)  per  square  centimeter,  to  the  square  of  the  frequency 
(/)  in  cycles  per  second,  and  to  the  mass  in  cubic  centimeters. 

Peozf2VB2, 

(38)  Pe  =  Kf2VB2. 

K  is  determined  by  the  formula  (39)  where  t  is  the  thick- 
ness of  the  material  in  centimeters  and  c  is  the  electric 
conductivity  in  mhos.  For  iron  c  =  100000  mhos  and  for 
copper  c  =  700,000  mhos. 

(39)  k  =  ~  X  10~9  t2c  - 1 .645  X  lO"9^. 

Ex.  24.  Substitute  the  value  of  K  from  (39)  in  (38),  and 
interpret  the  resulting  equation,  (b)  How  will  the  resulting  equation 
alter  when  Pe  is  expressed  in  watts,  V  in  cubic  feet,  t  in  inches, 
and  (#max)  in  lines  per  square  inch?  (c)  How  will  the  last  equa- 
tion read  when  the  thickness  is  expressed  in  mils? 

Ex.  25.  Calculate  the  eddy  current  loss  for  the  core  of  a 
generator  giving  60  cycles  per  second.  The  volume  of  iron  is 
11  cu.ft.    The  thickness  of  lamina  =0.1  in. 

Ex.  26.  A  magnetic  circuit  consists  of  a  wrought-iron  bent 
bar,  250  cm.  length  and  50  sq.cm.  in  cross-section,  with  an  air- 
gap  of  .5  cm.  length.  How  many  ampere-turns  are  necessary 
to  set  up  (a)  50000  lines  of  force,  (b)  100000  lines,  (c)  500000  lines, 
(d)  1000000  lines? 

Ex.  27.  A  magnetic  circuit  consists  of  150  cm.  of  cast  steel, 
with  60  sq.cm.  in  cross-section;  135  cm.  of  sheet  steel,  with  55 
sq.cm.  cross-section;  and  1  cm.  of  air  with  55  sq.cm.  of  cross- 
section.  There  are  1500  turns  of  wire  wound  upon  this  circuit. 
How  many  amperes  will  be  necessary  to  set  up  600000  lines? 


THE  ALGEBEA  OF  THE  MAGNETIC  CIRCUIT     193 

Ex.  28.  Construct  a  table  of  hysteresis  loss  for  transformers 
used  on  25-,  60-,  and  133-cycle  circuits.  The  range  of  core  weights 
to  extend  from  50  to  300  lbs.  in  gradations  of  25  lbs. 

16.  The  average  E.M.F.  which  is  induced  in  an  inductor, 
i.e.,  a  moving  wire  or  electric  conductor,  is  proportional 
to  the  flux  density  (H),  and  to  the  length  (I)  and  velocity 
(v)  of  the  inductor, 

E.M.F.  ex  Hlv. 

In  the  C.G.S.  system  the  proportionality  factor  is  one 
and  therefore, 

(40)  EM.F.  =  Hlv. 

The  average  E.M.F.  is  equal  to  the  total  flux  (<£)  divided 
by  the  time  if)  in  which  it  is  cut. 

(41)  E.M.F. =y (abvolts)  =^  volts. 

Transform  and  interpret  equations  (40)  and  (41). 

Observation.  One  volt  is  induced  in  an  inductor  by  the 
cutting  of  100000000  lines  of  force  per  second.  One  inductor 
of  an  armature  cuts  2$  lines  of  force  in  one  revolution  for 
each  pair  of  poles. 

17.  The  force  tending  to  push  a  conductor  aside  in  a 
magnetic  field  varies  as  the  product  of  the  strength  of  the 
field,  the  length  of  the  conductor  and  the  current  flowing 
through  the  conductor. 

(42)  F  =  IIH  dynes. 

Interpret  this  equation  for  a  generator  and  for  a  motor. 

18.  The  E.M.F.  generated  in  any  direct  current  arma- 
ture is  directly  proportional  to  the  product  of  the  total 
flux  ($)  which  is  cut,  the  total  number  (N)  of  conductors 
on  the  periphery  of  the  armature,  the  number  (p)  of  poles 
and  the  number  of  revolutions  per  second  (n)  and  inversely 
proportional   to   the   number    (q)    of  parallel   paths.     The 


194  PRACTICAL  MATHEMATICS 

proportionality  factor  is  10-8  when  the  result  is  expressed 
in  volts. 

(43)  E.M.F.  =  ~|?  (volts). 

Ex.  29.  In  the  air-gap  of  a  generator  a  flux  of  2000000  maxwells 
exists  under  each  pole.  At  a  speed  of  1200  R.P.M.  what  E.M.F. 
is  generated  in  each  conductor  of  the  armature? 

Ex.30.  If  the  poles  cover  70  per  cent  of  the  armature  surface 
of  the  above  gererator  and  the  armature  is  1  ft.  long,  1  ft.  in 
diameter  and  the  field  strength  5000  gausses  at  the  armature 
surface,  what  average  E.M.F.  is  generated  in  each  conductor? 

19.  The  work  (W)  done  by  the  armature  of  a  generator 
or  motor  may  be  expressed  electrically  by  (44)  and  mechanic- 
ally by  (45). 

(44)  W  =  EI  watts. 

where  m  =  number  revolutions  per  minute, 
T  =  torque  in  foot-pounds, 
Equate  (44)  and  (45)  and  obtain  (46). 

Tjjmm{tAhs 

m 

In  (46)  substitute  the  value  of  E  from  (43)  and  interpret. 

Torque  is  the  product  of  the  peripheral  force  (F)  of  the 
armature  multiplied  by  the  mean  R  radius  of  the  armature 
winding. 

(47)  T  =  FR=—~. 

Solve  for  F  and  R  and  interpret. 

The  torque  or  turning  moment  necessary  to  cause  a 
generator  armature  to  rotate,  and  the  torque  produced  by 
a  motor  armature  are  of  precisely  the  same  nature.     Each 


THE  ALGEBRA  OF  THE  MAGNETIC  CIRCUIT     195 

depends  upon  the  strength  of  the  field,  the  armature  current, 
the  length  of  coil  and  its  displacement  from  the  axis  of 
rotation. 

The  peripheral  force  divided  by  the  number  of  effective 
conductors  (Nc)  gives  the  peripheral  force  (Fc)  acting  on 
each  armature  conductor. 

(48)  F..=|,      . 

Ex.  31.  In  the  two  preceding  problems  (29),  (30),  calculate 
the  force  in  pounds  upon  each  conductor  tending  to  resist  the 
motion  of  the  armature  when  the  current  in  each  conductor  is 
25  amps. 

20.  Flux  Density  in  Armature.  For  slotted  cores  the 
flux  density  will  be  greater  in  the  teeth. 

(AQ)  ^-  =  ?i= l^L 

k    7  *a      Ba       bt  +  bs+B~f+btf 

Interpret  (49),  simplify  (49)  and  reinterpret. 

i?a  =  the  apparent  flux  density  in  the  teeth  in  lines 

per  square  inch, 
Bt  =  the  actual  flux  density  in  the  teeth  in  lines  per 

square  inch, 
bt  =the  mean  width  of  teeth  in  inches, 
bs  =  the  mean  width  of  slot  in  inches, 
(i=the  permeability  of  the  tooth  corresponding  to 

Bt, 
f=the  ratio  of  the  net  length  to  the  gross  length  of 
the  armature  core. 

Ex.  32.  The  total  ampere-turns  in  any  D.C.  armature  winding 
equals  one-half  the  product  of  the  number  of  conductors  times 
the  current  per  conductor. 

Write  this  as  a  formula. 

Ex.  33.  Compute  the  force  acting  on  a  conductor  of  25  cm. 
length.  The  current  passing  through  the  conductor  is  5  abamps 
and  the  field  strength  is  3000  gausses. 


196  PRACTICAL  MATHEMATICS 

Ex.  34.  A  circular  wire  of  50  cm.  radius  rotates  25  times  per 
second  in  a  magnetic  field  of  1000  gausses.  What  voltage  is 
generated  in  the  wire? 

Ex.  35.  Determine  the  E.M.F.  of  a  10-pole  generator  making 
25  revolutions  per  second.     The  flux  equals  2500000. 

Ex.  36.  A  2-pole  machine  has  a  flux  of  3000000  lines  per 
pole.  There  are  200  conductors  and  2  parallel  paths.  The 
machine  rotates  at  20  revolutions  per  second.  What  voltage  is 
developed? 

Ex.  37.  A  110-volt  motor  of  a  motor-generator  set  takes 
100  amperes  and  transmits  with  15  per  cent  loss  to  a  generator 
of  84  per  cent  efficiency.  The  generator  shows  20  volts.  What 
current  is  it  giving  off? 

Ex.  38.  A  compound  generator  at  full  load,  which  is  50  amperes 
at  110  volts,  requires  11000  ampere-turns  and  at  no  load  8500 
ampere-turns.    How  many  turns  are  there  in  the  series  field? 


CHAPTER  X 


WINDING  CALCULATIONS 


1.  The  dimensions  of  an  insulated  wire  are,  d  the  diam- 
eter of  the  bare  wire,  and  d\  the  outside  diameter  over 
insulation  and  lead  casing.  Designate  the  thickness  of 
insulation  by  t,  then  d\  =  d+2t. 

1  =  length  and  6  =  depth  of  cross-section  of  space  to  be 
filled  with  wire. 

2.  Square  Winding.  Assume  I  and  b  to  be  exact  multi- 
ples of  d\.    The  space  may  be  considered  as  divided  into 


Fig.  63. — Square  Winding. 

squares.  Each  square  will  circumscribe  a  circle.  The 
circles  will  be  tangential  and  their  centers  will  be  in  hori- 
zontal and  vertical  alignment  as  in  Fig.  63.  The  number  of 
turns  of  wire  is  given  in  (1). 


(1) 


N 


lb 


How  many  wires  are  there  per  layer?    How  many  layers 
are  there? 

The  ratio  of  the  cross-section  of  metal  to  the  square 
space  which  is  required  is  called  the  space  factor  <j. 

197 


198 


PRACTICAL  MATHEMATICS 


Ex.  1.    Show  that  for  square  winding 
(2)  ,=0.7854^. 

Ex.  2.     Show   that  the   total    cross-section   of   wire   is 


Iba 


Simplify  by  substituting  for  <j  from  (2). 

3.  Stagger  or  Imbedded  Winding.  The  space  may  be 
considered  as  divided  into  hexagons.  Each  hexagon  will 
circumscribe  a  circle.  The  circles  will  be  tangential.  Their 
center  lines  will  be  horizontal  and  will  also  incline  60°  out 
of  vertical  alignment. 

Ex.  3.  In  Fig.  64  determine  the  value  of  q,  which  is  the  vertical 
distance  between  horizontal  center  lines. 


Fig.  64. — Imbedded  Winding. 

Ex.  4.    In  this  example  consider  I  an  exact  multiple  of  dx. 

I  I 

Then  in  one  row  there  will  be  -  circles,  suppose  we  say  -r  =x. 

How  many  circles  will  there  be  in  two  rows? 

How  many  circles  will  there  be  in  three  rows? 

How  many  circles  will  there  be  in  four  rows? 

How  many  circles  will  there  be  in  n  rows? 

What  is  the  depth  of  the  rectangle  enclosing  one  row  of  circles? 

What  is  the  depth  of  the  rectangle  enclosing  two  rows  of  circles? 


WINDING  CALCULATIONS  199 

What  is  the  depth  of  the  rectangle  enclosing  three  rows  of 
circles? 

What  is  the  depth  of  the  rectangle  enclosing  n  rows  of  circles? 

Ex.  5.     (a)  What  is  the  metallic  cross-section  for  n  rows  of 
wire  of  outside  diameter  dh  having  x  wires  in  the  lower  row? 
(6)  What  is  the  area  of  the  enclosing  rectangle  for  n  rows? 

(c)  Determine  a  where  a  =-r-. 

(&) 

(d)  Calculate  a  for  one,  two,  three,  four  layers  of  winding. 

Ex.  6.  Determine  a  for  a  winding  of  rectangular  bar  of  dimen- 
sions a  and  b  and  insulation  thickness  t  where  the  winding  space 
is  filled  by  the  material. 

The  values  of  <j  in  Ex.  1-5  are  for  ideal  conditions  which 
can  be  only  approximated  in  practice.  The  value  of  <j 
is  more  often  expressed  in  terms  of  d,  i.e.,  the  bare  diameter 
of  the  wire. 

4.  Number  of  Turns. 

(3)  N-!*, 

where  i4c  =  area    of    one   conductor   (bare)   in   centimeters. 
Solve  for  vlbAc  and  interpret. 

5.  Diameter  of  Wire  in  Terms  of  Resistance  and  Wind- 
ing Volume. 

(4)  W  =  PR  represents  the  permissible  loss  in  watts. 

{b)  K~d2d12' 

V  =  volume  of  winding  space, 
K  =  sl  constant  varying  with   allowance  rise  in  the 

temperature  of  coil, 
K  =  0.8484  for  15°  C.  rise, 
#  =  0.9001  for  30°  C.  rise, 
K  =  1.405  for  60°  C.  rise. 


200 


PRACTICAL  MATHEMATICS 


Solve  (5)  for  d  and  substitute  for  R  from  (4)  simplify 
and  interpret  these  equations. 

Ex.  7.  A  spool  of  wire  is  rewound  with  wire  having  twice  the 
number  of  circular  mils.  Select  any  convenient  size  for  the  first 
wire.    What  will  be  the  relative  resistance  of  the  two  windings? 

6.  Volumes  of  Winding  Space.     The  volume  (V)  of  any 

winding  space  is  determined  by  multiplying  the  transverse 

sectional  area  of  the  winding  space  by  its  length  (I).     The 

x 
area  of  an  ellipse  equals  —  times  the  product  of  its  long  and 

short  diameters. 

Ex.  8.  Verify  and  interpret  the  following  formulas  for  the 
volumes  of  the  winding  space  corresponding  to  the  sections  shown 
in  Fig.  65. 


CO     ,o-> 

a 

- 

Fig.  65. — Winding  Space  Sections. 


(6) 

V  **&***% 

Circular  core 

(7) 

V-l(A*-a*), 

Square  core 

(8) 

y  =  j{^-(A'-a!)+m(A-a)} 
k  4 j                                      ' 

Link  core 

(9) 

V^iAB-ab), 

Elliptic  core 

(10) 

V  =  b[(l+%)(A-a)+2a}. 

Rectangular  core 

(11) 


7.  The  heating  of  magnets  is  given  by  (11), 

P 


T  =  K- 


WINDING  CALCULATIONS  201 

Interpret  (11)  according  to  the  following  notation: 

27  =  use  in  temperature  centigrade, 

P  =  power  in  watts  dissipated  in  the  coil, 

A  =  outside  cylindrical  surface  of    he  coil  in  square 

inches, 
K  =  temperature  rise  in  centigrade  per  watt  per  square 

inch  of  outside  cylindrical  surface, 
K=  130  for  open  electromagnets, 
K  =  95  for  iron-clad  electromagnets, 
K  =  70  for  field  magnets  open, 
K  =  140  for  field  magnets  closed. 

Ex.  9.    Determine  the  rise  in  temperature  of  the  magnet 
specified  in  Ex.  11,  Chapter  IX. 


CHAPTER  XI 
FORMULAS  OF  MENSURATION 

1.  The  formulas  of  this  chapter  are  the  familiar  formulas 
of  mensuration  supplemented  by  practical  formulas  for 
appr  oxim  ations . 

The  area  of  any  figure  is  its  ratio  to  a  unit  of  area.  The 
area  of  a  surface  or  a  cross-section  of  a  solid  may  be  measured 
directly  or  indirectly  by  comparing  it  with  any  shaped 
surface  or  cross-section.  The  unit  of  area  takes  its  name 
from  the  shape  of  the  unit  figure.  Thus  unit  areas  which 
are  square  are  called  square  units.  A  unit  area  which  is 
a  circle  is  called  a  circular  unit.  A  unit  area  which  is  a 
triangle  is  a  triangular  unit.  All  such  units  of  area  have 
their  linear  dimensions  equal  to  one  unit.  A  square  unit 
area  having  a  side  equal  to  1  mil  is  a  square  mil.  A  cir- 
cular unit  area  having  a  diameter  of  1  mil  is  a  circular  mil. 

The  volume  of  a  figure  is  its  ratio  to  a  unit  of  volume. 
The  volume  of  a  solid  or  geometric  figure  of  three  dimensions 
may  be  measured  directly  or  indirectly  by  comparing  it 
with  any  shaped  solid  or  geometric  figure  of  three  dimen- 
sions. The  unit  of  volume  takes  its  name  from  the  shape 
of  the  unit  figure.  Thus  unit  volumes  which  are  cubes 
are  called  cubic  units.  A  unit  volume  which  is  a  sphere  is 
called  a  spheric  unit.  A  unit  volume  which  is  a  cylinder, 
1  foot  in  length  and  1  circular  mil  in  cross-section  is  called 
a  circular  mil  foot.  A  board  foot  is  a  rectangular  prism,  1 
ft.  in  length  and  having  a  rectangular  cross-section  of  12 
sq.ins.  A  cube  having  its  edges  equal  to  1  cm.  is  a  cubic 
centimeter. 

»  202 


FOEMULAS  OF  MENSURATION 


203 


2.  Area  of  a  Parallelogram.  The  area  (A)  of  a  parallel- 
ogram equals  the  product  of  its  base  (6)  times  its  altitude 
(h).  Formulate  this  law  and  solve  for  b  and  h  and  interpret 
the  resulting  equations. 

In  Fig.  66  K  is  the    diagonal,  a  the  angle    opposite  the 


Fig.  66. — A  Parallelogram. 

diagonal  and  c  the  side  adjacent  to  the  base.     Express  the 
area,  altitude  and  the  diagonal  in  terms  of  b,  c  and  a. 

How  do  these  various  formulas  simplify  when  the  figure 
becomes  a  rectangle  and  also  when  it  becomes  a  square? 

What  is  the  relation  of  the  opposite  angles  of  the  parallel- 
ogram?    What  is  the  sum  of  the  angles  of  the  parallelogram? 

3.  Area  of  a  Triangle.  What  relation  does  a  triangle 
bear  to  a  parallelogram  which  has  an  equal  base  and  an 
equal  altitude.  Formulate  the  law  for  the  area  of  a  triangle 
in  terms  of  b  and  h,  solve  and  interpret  the 
resulting  equation  for  b  and  h. 

In  Fig.  67  a  is  the  angle  between  the  base 
b  and  the  adjacent  side  c.  Express  the  area 
and  the  altitude  of  the  triangle  in  terms  of  b, 
c  and  a. 

How  do  these  formulas  simplify  when  the 
figure  becomes  a  right  triangle,   an  isosceles  triangle,  and 
an  equilateral  triangle? 

The  general  formula  for  the  area  of  a  triangle  is  given 
in  (1)  in  which  6,  c  and  c'  are  the  three  respective  sides  and 
5  is  an  abbreviation  for  half  the  sum  of  the  sides. 


Fig.  67. 


(1) 
(2) 


A  =  Vs(s-b)(s-c)(s-c') 
b+c+cf 


204  PRACTICAL  MATHEMATICS 

Substitute  —for  A  and  solve  for  b  and  h. 
2d 

4.  Area  of  a  Trapezoid.  Construct  a  trapezoid,  desig- 
nating its  parallel  sides,  i.e.,  bases  by  (6)  and  (e)  and  the 
altitude  by  (h),  the  diagonals  by  (&i)  and  (fo)  and  the  angles 
opposite  the  diagonals  by  (a)  and  (g)  respectively. 

Write  all  the  possible  formulas  for  the  area  and  the 
diagonals  in  terms  of  the  mentioned  parts. 

5.  Area  of  a  Regular  Polygon.  What  is  the  magnitude 
of  the  central  angle  which  is  subtended  by  half  the  side? 

In  Fig.  68  the  side  is  designated  by  b, 
the  apothem  by  a  and  the  radius  of  the 
circumscribing  circle  by  r. 

A  regular  polygon  is  divided  into  n 
equal  isosceles  triangles  (/)  and  therefore 
its  area  is  computed  by  multiplying  the 

Pjq  68 A  Regu-     area  °f    CO    by   n,   which    is    also   the 

lar  Polygon.  number  of  sides.     Formulate  the  law  for 

the  area  of  a  polygon  in  terms  of  a,  b, 
and  n,  also  in  terms  of  b,  r,  n,  and  the  central  angle  sub- 
tended by  the  half  side.  In  the  above  equations  substitute 
(p)  the  perimeter,  for  bn  and  interpret  the  resulting 
equations. 

6.  Area  of  a  Circle.  The  circumference  (c)  of  a  circle 
equals  x  times  its  diameter  (d).  The  value  of  x  is  an  unend- 
ing decimal  and  is  approximately  represented  by  3.1416 

and  by  — . 

The  area  of  a  circle  equals  one-half  the  product  of  its 
circumference  times  its  radius  (r).  Formulate  the  law  for 
the  area  of  a  circle.  Substitute  for  c  in  terms  of  d  and  solve 
for  d  and  interpret.  Substitute  for  d  in  terms  of  r  and  inter- 
pret the  equations  of  the  area  and  of  the  circumference 
in  terms  of  r. 

A  central  angle  of  1°  intercepts  an  arc  equal  to  3+ff  of 
the  circumference.     What  part  of  the  circumference  will 


FORMULAS  OF  MENSURATION 


205 


be  intercepted  by  a  central  angle  of  0°?     A  central  angle 

57.3      1 
of  57.3°   (approximate)  will  intercept  an  arc  of  — —  *=*  — 

360      2% 

of  the  circumference.  The  central  angle  of  57.3°  is  called 
a  radian  of  angle  and  its  intercepted  arc  is  called  a  radian 
of  arc.  Show  that  a  radian  of  arc  equals  the  radius  of  the 
circle?  The  length  of  an  arc  (L)  equals  the  radius  times 
the  number  of  radians  (0)  in  the  arc.     L  =  rQ. 

7.  Area  of  an  Annulus  or  Circular  Ring.  An  annulus 
is  the  figure  formed  by  a  circle  interior  to  another  circle. 
The  area  of  an  annulus  is  the  difference  in  areas  between 
the  outer  and  inner  circle.  In  Fig.  69  d\  and  cfe  are  the 
respective  diameters.  Formulate  the  law  for  the  area  of  an 
annulus  in  terms  of  the  diameters  and  also  in  terms  of  the 
radii. 


Fig.  69. — An  Annulus. 


Fig.  70.— An  Ellipse. 


8.  Area  of  an  Ellipse.  The  area  of  an  ellipse  equals 
.7854  times  the  product  of  the  major  diameter  (di)  times 
the  minor  diameter  (^2).  Formulate  the  law  for  the  area 
of  an  ellipse. 

In  Fig.  70  half  the  major  and  minor  diameters,  i.e., 
the  semimajor  and  semiminor  diameters,  are  designated 
by  (a)  and  (b)  respectively.  Formulate  the  area  of  an 
ellipse  in  terms  of  a  and  b. 

9.  "  The  squaring  of  the  circle '"  is  a  problem  handed 
down  from  the  classic  days  of  Greece.  It  was  an  attempt 
to  determine  by  the  aid  of  the  straightedge  and  compass 
a  square  exactly  equal  in  area  to  a  circle.     Until  the  sig- 


206  PKACTICAL  MATHEMATICS 

nificance   of  ir  was   appreciated   this   problem  remained 
unsettled. 

The  length  of  an  arc  of  a  circle  is  given  in  (3),  where 
L  =  length  of  the  arc,  c\  represents  the  length  of  the  sub- 
tended chord,  and  C2  the  length  of  the  chord  which  is 
subtended  by  half  the  arc.  This  is  known  as  Huyghen's 
approximation. 

(3)  L=^p. 

Interpret  (3). 

Another  approximate  formula  is  given  in  (4)  where 
C4  is  the  length  of  the  chord  subtended  by  one-fourth  of 
the  arc. 

(4)  L  =  ^(ci-40c2+256c4). 
Interpret  (4). 

Ex.  1.  Construct  arcs  of  30°,  60°,  90°,  120°,  150°,  180°,  and 
270°  with  a  radius  of  10  ins.  Measure  the  values  of  ch  c2,  and 
d  and  substitute  in  (3)  and  (4).  Compare  the  results  of  (3)  and 
(4)  by  computing  the  arcs  in  terms  of  their  respective  central 
angles.    See  Table  VIII  for  the  values  of  chords  for  unit  circles. 

10.  The  rectification  of  a  circular  arc  is  the  determina- 
tion of  the  length  of  a  straight  line  which  shall  be  equal 
to  the  given  arc.  In  Fig.  71  the  rectification  of  the  semi- 
circle AVB  is  represented  by  FS. 

The  construction  suggested  by  Ceradini  is  as  follows: 
OF  is  a  radius  constructed  perpendicular  to  the  hori- 
zontal diameter  AB;  the  tangents  VC  and  BC,  drawn 
from  V  and  B  respectively,  intersect  at  C;  the  tangent 
AF  is  constructed  equal  to  A B  and  terminates  at  F. 
With  V  as  a  center  strike  an  arc  equal  to  the  radius  OV 
and  intersecting  the  circumference  at  D;  draw  the  secant 
VS  through  V  and  D  intersecting  CB  at  S;  join  S  with 
F.  A  proportional  part  of  the  semicircumference  will 
be  obtained  by  dividing  FS  proportionately. 


FORMULAS  OF  MENSURATION 


207 


A  second  method  of  rectification  given  by  Rankine 
is  presented  in    Fig.   72.      An    arc   AV  with    center  0 


Fig.  71. — Rectification  of  ?.n  Arc. 


Fig.  72. — Rankine's  Method. 


is  to  be  rectified.     Draw  the  chord  AV  and  bisect  it. 
Extend  A  V  to  T,  making  AT  =  §AV.    Draw  CV  tangent 


208 


PEACTICAL  MATHEMATICS 


to  the  arc  and  therefore  perpendicular  to  OF  at  V.     With 

T  as  a  center  strike  the  arc  AC.     Then  arc  AV=CV. 
A  third  method  of  rectifying  an  arc  AC  which  was 

suggested  by  Snell  is  as  follows:  Construct  the  semi- 
circle ACB  with  diameter  A B,  and 
the  tangent  AE  at  A.  Extend  the 
diameter  through  B  to  D  making 
BD  =  %  AB.  A  secant  passing 
through  D  and  C  cuts  the  tangent 
AE   at   E.      Then   AE  =  arc   AC. 

The   secant   cuts   the  semicircle  at  the  second  point  C . 


Fig.  73  — Snell's  Method. 


the  approximation  is  slightly  in  excess  whereas  the  former 
is  slightly  deficient. 

It  may  be  necessary  to  determine  the  length  of  the 
arc  of  a  circle  which  shall  be  equal  to  a  given  line.  The 
method  is  presented  in  Fig. 
74.  The  given  line  BX  is 
a  portion  of  a  tangent  to 
the  circle  at  B,  the  point 
which  marks  the  beginning 
of  the  equal  arc.  Locate 
L  so  that  BL  =  \BX.  From 
L  as  a  center  strike  an  arc 
of  radius  LX  intersecting 
the  circle  at  A.  Then  arc 
AB  =  BX. 

11.  The  approximate  per- 
imeter of  an  ellipse  is  given 
in  terms  of  the  semimajor 
diameter  (a)  and  semiminor 
diameter  (b)  in  equations  (5),  (6)  and  (7). 
(5)  L=x(a+6). 


Fig.  74. — Circularization  of  a 
Line. 


(6) 
(7) 


L=% 


L=|(a+6+V^+P). 


FORMULAS  OF  MENSURATION 


209 


Transform  (5)  and  (6)  and  solve  for  a  and  b. 

(5)  will  give  an  answer  .5  per  cent  too  small  and  (6) 
will  give  an  answer  .5  per  cent  too  large,  whereas  (7) 
will  give  an  answer  within  ^-J-^  per  cent  of  the  truth.  Sub- 
stitute a  =  4,  6  =  3  in  (5),  (6)  and  (7). 

12.  The  sector  of  a  circle  is  the  area  bounded  by  two 
radii  and  an  arc.  Express  the 
area  of  sector  AOD  of  Fig.  75,  in 
which  b  represents  the  arc  ACD 
and  r  represents  the  radius.  The 
area  of  a  sector  equals  one-half 
the  product  of  the  radius  times 
the  arc.  Substitute  the  value  of  b 
in  terms  of  the  radius  and  the  cen- 
tral angle  4>  according  to  5. 

Interpret    the    following    rela- 
tions : 


Fig. 


Area  of  sector 
Area  of  circle 


360 


arc  b 


circumference  2xr* 


13.  A  segment  of  a  circle  is  the  area  bounded  by  a 
chord  and  its  intercepted  arc.  The  sector  EGF  of  Fig 
75  has  a  chord  EF  designated  by  c  and  a  height  or  rise 
designated  by  h.  Express  the  area  of  segment  EGF  by 
subtracting  the  area  of  the  triangle  EOF  from  the  area 
of  the  sector  EOF.  Show  that  the  length  of  the  chord 
EF  is  expressed  by  (8). 


(8) 


<?  =  4h(2r-h). 


Show  that   (9)   expresses  the  length  of  the  chord  a 
Fig.  75,  which  subtends  the  central  angle  8. 


(9) 


a  =  2r  sin 


210  •       PKACTICAL  MATHEMATICS 

The  approximate  formula  for  the  area  of  a  segment 
is  given  in  (10)  and  (11),  and  the  exact  formula  in  (12). 

(10)  A  =  2c+-S' 

(11)  a  =  i^W+W). 

(12)  A=^(6-sin6). 

Ex.  2.  Compare  the  results  of  (10)  and  (11)  with  (12),  for 
a  circle  of  10  ins.  diameter,  in  which  the  chord  subtends  a  central 
angle  of  60°.    Transform  (11)  and  solve  for  c. 

14.  The  area  of  the  segment  of  an  ellipse  whose  chord 
c  is  perpendicular  to  one  of  its  diameters  (d)  equals  the 
product  of  its  major  and  minor  diameters  times  the  area 
(Ac)  of  the  segment  of  a  circle  of  diameter  (d)  and  whose 
chord  equals  c.     Formulate  this  law. 

Ex.  3.  Show  that  (13)  expresses  the  diameter  (d)  of  a  circle 
whose  area  equals  that  of  an  ellipse  of  diameter  di  and  ck. 

(13)  d  =  Vdrf~2. 

Interpret  (13),  transform  and  solve  for  d\  and  d2. 

15.  The  area  of  the  segment  of  a  parabola   Fig.  76, 

equals    two-thirds    the    area   of   the 
•* a *        circumscribing    rectangle.       Formu- 
late   this    law    in    terms    of   h   the 
j        height  of    the   parabola,  and  a  the 
/        chord  of  the  parabola.     The  height 
/   I        of    the   parabola   is    called   its  ordi- 

1       nate  and   half  the   chord   is    called 

Fig.  76.— A  Parabola.     its  abscissa.     The  mid  point  of  the 
base  of  the  rectangle   is    tangent  to 
the  parabola  at  a  point  called  its  vertex. 


FORMULAS  OF  MENSURATION  211 

16.  The  approximate   length  of   the  parabola  is  given 
by  (14). 


(i4)  *-*n|+T- 

Interpret  (14),  transform  and  solve  for  a  and  h. 

17.  The  approximate  area  A  of  the  segment  of  an 
hyperbola  is  given  in  (15)  and  the  approximate  length 
(L)  which  is  measured  from  the  vertex  is  given  in  (16). 

The  semimajor  and  semiminor  diameters  of  the 
hyperbola  are  represented  by  a  and  b  respectively  and 
the  ordinate  and  abscissa  of  the  curve  are  represented 
by  y  and  x  respectively.     See  Fig.  135. 

(15)  A  =  .16-z(V49a£+35x2+!vW). 

(16)  L  =  K1  +  (1.5a+2aiJ&*+.9^x)- 

Transform  and  solve  for  y  and  x. 

Simplify  (15)  and  (16)  by  factoring  and  by  the  appli- 
cation of  the  axiom  of  fractions. 

18.  Area  of  an  Irregular  Figure.  A  figure  bounded  by 
straight  lines  may  be  decomposed  into  triangles,  rectangles, 
and  trapezoids.  The  sum  of  the  areas  of  the  simpler  con- 
stituent figures  equals  the  area  of  the  entire  figure. 

When  the  boundaries  include  arcs  of  circles  the  figure 
will  include  sectors  and  segments. 

19.  When  the  outline  is  irregular  or  bounded  by  curves 
the  area  may  be  determined  by  one  of  the  following 
methods : 

(a)  by  using  a  planimeter, 
(&)  by  using  squared  paper, 

(c)  by  weighing, 

(d)  by  mean  ordinate  rule. 


212  PKACTICAL  MATHEMATICS 

Very  close  approximations  may  be  obtained  by  other 
methods  more  commonly  known  as: 

(e)  mid  ordinate  rule, 

(/)  trapezoidal  rule, 

(g)  Simpson's  one-third  rule, 

(h)  Simpson's  three-eighths  rule, 

(i)  Durand's  rule, 

0)  Weddle's  rule. 
(a)  The  Planimeter.     The  planimeter  is  an  instrument 
which  enables  us  to  measure  the  area  of  a  figure  in  square 
inches  or  any  other  units  of  area.     It  consists  primarily 
of  two  rigid  arms  AB  and  BC  pivoted  at  B. 


Fig.  77. — A  Planimeter. 

The  extremity  A  of  arm  AB  is  fixed  by  a  needle  point 
to  the  table.  The  tracing  point  C  of  arm  BC  is  carried 
around  the  outline  of  the  figure.  This  motion  causes 
rotation  of  the  wheel  W  which  is  graduated  on  its  rim. 
The  difference  of  the  readings  of  the  wheel  before  and  after 
using  the  instrument,  when  multiplied  by  a  suitable  constant 
gives  the  area  of  the  figure.  The  constant  (K)  of  the  instru- 
ment may  be  determined  by  tracing  the  outline  of  known 
area  A.  If  the  wheel  readings  before  and  after  tracing 
are  d\  and  cfe  respectively  then, 

The  horse-power  of  an  engine  may  be  determined  from 
the  indicator  diagram  which  shows  the  performance  of  an 


FORMULAS  OF  MENSURATION 


213 


engine  during  a  single  stroke.  The  vertical  scale  of  the 
diagram  in  pounds  per  inch  is  the  calibrated  test  number 
of  the  spring  used  in  the  indicator.  The  horizontal  scale 
of  the  diagram  per  inch  is  the  ratio  of  the  length  of  the 
stroke  in  feet  to  the  length  of  the  card  in  inches.  What 
is  the  value  in  foot-pounds  of  each  square  inch  of  the  indica- 
tor diagram.  The  total  area  of  the  indicator  diagram 
multiplied  by  the  effective  area  of  the  piston  in  square  inches 
times  the  number  of  strokes  per  minute  divided  by  33000 
gives  the  horse-power  of  the  engine.  The  planimeter  can 
be  used  to  determine  the  area  of  the  indicator  diagram. 

(b)  Use  of  Squared  Paper.  The  figure  may  be  drawn 
upon  cross-section  paper.  This  is  usually  ruled  horizontally 
and  vertically  with  10  or  20  divisions  per  square  inch.  The 
number  of  full  squares  enclosed  by  the  figure  is  counted 
and  with  a  little  care  and  practice  the  remaining  fractional 
squares  can  be  closely  approximated. 

(c)  Weighing  of  Areas.  After  the  figure  has  been  traced 
upon  heavy  cardboard  or  thin  sheet  metal  of  uniform 
thickness  its  outline  is  cut.  The  figure  is  then  weighed  and 
its  weight  compared  to  the  weight  of  similar  material  of 
known  area. 

(d)  Mean  Ordinate  Rule.  The  area  of  the  irregular, 
Fig.  78,  is  equivalent    to  a  rectangle,  whose   base    is  AB 


r 7T 

— ;- — i--p 

i 

-1 ,~x 

I             \ 
I            i 

i  /  \     i 

A£— 1 l_ 

\! 

i  V 

-1 — r — f- 

i 

._} — j___j — j — i- 

j         A 

1    V 

i       V 
i _s 

-i 

i 
i 

i 

— I— 

III                             | 

| 
I       ^ * 

Fig.  78. — Area  by  Mean  Ordinate  Rule. 


which  is  also  the  length  of  the  figure,  and  whose  altitude 
CD  is  the  mean  of  all  the  ordinates,  i.e.,  perpendiculars 


214 


PRACTICAL  MATHEMATICS 


drawn  to  AB  within  the  figure.  The  mean  of  the  ordinates 
extending  above  AB  is  CB  whereas  the  mean  of  the  ordinates 
extending  below  AB  is  BD.  The  product  of  the  mean 
ordinate  CD  of  the  whole  figure  multiplied  by  its  length 
AB  is  the  area  of  the  figure. 

How  may  we  determine  the  mean  ordinate  of  a  figure 
from  a  knowledge  of  the  value  of  its  area. 

The  mean  effective  pressure  of  an  engine  is  determined 
from  its  indicator  diagram,  by  dividing  the  area  of  the 
diagram  by  its  length.  The  horse-power  of  an  engine  is 
given  by  (17),  in  which  P  is  the  mean  effective  pressure 
per  square  inch,  A  the  effective  area  of  the  piston  in  square 
inches,  L  the  length  of  the  stroke  in  feet,  and  N  the  number 
of  strokes  per  minute. 


(17) 


Horse-power  = 


PLAN 
33000  ' 


(e)  Mid  Ordinate  Rule.     The  base  AB  of  the  irregular 
Fig.  79,  is  divided    evenly  into    equal    spaces.     Ordinates 


Fig.  79. — Area  by  Mid  Ordinate  Rule. 


which  are  represented  by  dotted  lines  are  erected  at  the 
mid  points  of  the  spaces. 

The  sum  of  these  ordinates  divided  by  the  number 
erected  gives  an  approximation  to  the  mean  ordinate  of  the 
figure,  the  accuracy  of  which  increases  with  the  number 


FORMULAS  OF  MENSURATION 


215 


of  measured   ordinates.     It  is  sufficient  to  measure   only 
the  segments  of  the  o  dinates  which  lie  within  the  figure. 

(f)  Trapezoidal  Rule.  The  base  of  a  figure  is  divided 
into  any  number  of  equal  divisions.  In  Fig.  80,  the  base 
or  line  of  greatest  length  is  divided  into  twelve  equal 
divisions.  At  these  points  ordinates  are  erected  dividing 
the  figure  into  trapezoids  (approximate)  of  equal  width  S. 
The  area  of  each  trapezoid  equals  the  product  of  S  times 
the  half  sum  of  its  bounding  ordinates. 


y0  Vi  v*  y3  v*  y$  %  v,  y*  1/9  2/10  Vn  tfa 


yu  Vu  Vu  Via  Vn  Vi*  VM  Vso  %i  V-a  Vu  V*  fc 

Fig.  80. — Area  by  Tfapezoidal  Rule. 

The  ordinates  measured  above  the  base  are  designated 
by  2/o,  2/i,  2/2,  .  .  •  2/12  and  those  measured  below  the  base 
by  2/13,  2/i4,  .  .  •  2/25. 

Show  that  the  total  area  of  the  Fig.  80  equals  S  times 
the  sum  made  up  by  taking  one-half  the  outside  ordinates, 
to  which  is  added  the  sum  of  the  intermediate  ordinates. 
The  two  segments  of  an  ordinate  need  not  be  measured 
separately. 

For  a  figure  divided  into  seven  equal  parts  the  area  is 
expressed  in  (18). 

(18)  Area  =  /S{  |(2/o+2/6)+2/i+2/2+2/3+2/4+2/5} 

(g)  Simpson's  One-third  Rule.  A  closer  approximation 
for  the  area  than  can  be  obtained  by  the  trapezoidal  rule 
is  given  in  (19)  and  is  known  as  the  one-third  rule  for  an 


216  PRACTICAL  MATHEMATICS 

odd  number  of  ordinates  except  seven.  E  represents  the 
sum  of  the  end  ordinates;  B  represents  the  sum  of  the  even 
ordinates;  and  C  represents  the  sum  of  the  odd  ordinates. 

(19)  Area=|(#+4J3+2C). 

Show  that  (20)  expresses  the  area  of  a  figure  when  nine 
ordinates  are  used  and  designated  by  the  y  notation. 

S 

(20)  Area  =  -^{2/o+2/8+4(i/i+2/3+2/5+2/7)+2(2/2+2/4+2/6)!. 

For  the  special  ease  of  seven  ordinates  the  trapezoidal 
rule  is  more  accurate.  The  one-third  rule  is  most  accurate 
when  the  curve  approaches  a  parabola. 

(h)  Simpson's  Three-eighths  Rule.  For  an  even  number 
of  ordinates  a  close  approximation  is  given  in  (21).  E  repre- 
sents the  sum  of  the  end  ordinates;  M  represents  the  sum 
of  the  two  middle  ordinates;  R  represents  the  sum  of  the 
remaining  ordinates. 

(21)  Area  =  ^(#+2  M+3  R). 

o 

00  Durand's  Rule.  A  close  approximation  is  given  in 
(22).  E  represents  the  sum  of  the  end  ordinates;  P  repre- 
sents the  sum  of  the  *penultimates  (the  second  and  next  to 
the  last  ordinate) ;  R  represents  the  remaining  ordinates. 

'22)  Area  =  £{.4  E+l.l  P+R] 

(j)  Weddle's  Rule.  When  the  boundary  is  a  continuous 
curve  (23)  is  especially  applicable.  B  represents  the  sum 
Of  the  even  ordinates;  E  represents  the  sum  of  the  end 
ordinates;  C  represents  the  sum  of  the  odd  ordinates;  M 
represents  the  sum  of  the  middle  ordinates. 

(23)  Area=  .3  S(5 B+C+E+M). 

Ex.  4.  Determine  the  area  of  the  hysteresis  loop,  Fig.  190, 
by  the  different  methods  outlined  above.  Arrange  the  results 
in  the  order  of  their  magnitude. 


FORMULAS  OF  MENSURATION  217 

20.  The  Cycloid.  The  cycloid  is  a  curve  generated,  i.e., 
traced  by  a  point  on  a  circle  which  is  rolled  on  a  straight 
line  called  the  base.  The  length  of  the  base  equals  the  cir- 
cumference of  the  generating  circle.  The  length  of  the 
cycloid  equals  four  times  the  diameter  of  the  generating 
circle.  The  area  between  the  cycloid  and  the  base  equals 
three  times  the  area  of  the  generating  circle. 

21.  The  Helix.  The  helix  or  curve  of  a  screw  thread  is 
generated  by  advancing  and  rotating  a  point  on  a  cylinder. 
The  length  of  a  helix  is  given  in  (24) .  C  represents  the  cir- 
cumference of  the  cylinder;  P  represents  the  pitch  of  the 
screw  or  the  axal  advance  of  the  point  in  one  revolution; 
N  represents  the  number  of  advances  or  the  number  of  revo- 
lutions about  the  axis. 


(24)  Length  =  NVC2+P2. 

Transform  and  solve  for  JV,  C,  and  P. 

22.  The  Spiral.  The  spiral  is  a  curve  which  is  generated 
by  a  point  rotating  about  a  pole,  i.e.,  a  fixed  center,  so  that 
its  polar  distance  is  continuously  increased.  A  flat  spring 
is  a  plane  spiral  in  which  the  polar  distance  is  increased 
uniformly.  Its  length  is  given  in  (25)  in  which  Di  and  Z>2 
are  the  greater  and  lesser  diameters  and  N  the  number  of 
convolutions. 

(25)  Length  =  dv(^±^). 

Transform  and  solve  for  N,  D\  and  D2 

23.  The  Conical  Spiral.  When  the  spiral  motion  is 
traced  on  a  cone  it  is  called  a  conical  spiral.  The  length  of 
a  conical  spiral  is  given  in  (26)  in  which  Di  and  D2  are  the 
greater  and  lesser  diameters,  N  the  number  of  convolutions 
and  h  the  height  of  the  polar  axis. 


(26)  Length  =  ^Wpi±^)}2+R 

Transform  (26)  solving  for  N,  h,  D\  and  D2. 


218 


PRACTICAL  MATHEMATICS 


24.  The  volume  of  any  figure  is  its  ratio  to  a  unit  volume. 
The  unit  of  volume  may  be  any  figure  of  volume  but  is  gen- 
erally considered  as  a  cube.  The  twelve  edges  of  the  unit 
cube  are  1  unit  in  length.  The  six  faces  or  bounding  sur- 
faces are  1  sq.in.  in  area.  At  each  corner  there  are  three 
mutually  perpendicular  edges  which  extend  in  the  three 
distinct  directions  right-left,  up-down,  and  front-rear. 
The  dimensions  of  the  cube  are  the  lengths  of  three  edges 
measured  in  the  three  directions.  The  measure  of  the 
extension  of  a  figure  of  volume  in  these  three  directions 
are  variously  termed  the  length,  breadth,  thickness,  height, 
width  and  depth. 

In  Fig.  81,  /is  the  unit  cube.     The  bottom  surface  on 


Fig  81. — The  Cube  and  its  Extensions. 


which  the  cube  is  supposed  to  rest  is  called  its  lower  base 
to  distinguish  it  from  the  top  or  upper  base  which  is  parallel 
and  equal  to  it.  The  vertical  faces  which  include  the  front, 
rear,  and  the  two  end  surfaces  are  called  by  the  collective 
name  of  lateral  surface. 

If  the  length  of  an  edge  of  the  cube  is  one  inch,  the 
volume  of  the  cube  is  one  cubic  inch  (cu.in.).  If  the  length 
of  the  edge  of  a  cube  is  one  centimeter,  then  the  volume  of 


FORMULAS  OF  MENSURATION 


219 


the  cube  is  one  cubic  centimeter  (cu.cm.).     The  cubic  unit 
takes  its  name  from  the  linear  unit  of  its  edge. 

A  new  solid  called  a  rectangular  prism  is  formed  by- 
extending  any  one  of  the  three  sets  of  the  four  parallel 
edges  of  cube  /  Fig.  81,  providing  the  three  pairs  of  opposite 
faces  retain  their  parallelism.  The  vertical  extension 
produces  III,  which  has  a  base  1  unit  square  and  a  height 
of  /  units.  Ill  will  have  /  times  the  volume  of  I.  In  II 
the  extension  has  been  to  the  right  side  and  in  IV  the 
extension  has  been  to  the  rear.  The  volume  of  II  will 
equal  b  times  the  volume  of  I  and  the  volume  of  IV  will 
equal  e  times  the  volume  of  I. 


Fig.  82. — The  Rectangular  Prism. 

In  Fig.  82  we  have  a  rectangular  prism  built  by  extend- 
ing the  cube  to  the  rear. 

The  dimensions  of  the  cube  are  1X1X1,  and  the  dimen- 
sions of  the  prism  are  1X1X&.  The  volume  of  the  prism 
equals  b  times  the  volume  of  the  cube. 


Fig.  83. — Extensions  of  a  Prism. 


If  the  prism  of  Fig.  82  is  extended  in  the  vertical 
direction  as  shown  in  the  left  of  Fig.  83,  its  dimensions 
are  eXlXb,  and  the  latter  has  e  times  the  volume  of  the 


220 


PEACTICAL  MATHEMATICS 


prism  in  Fig.  82,  and  therefore  eb  times  the  volume  of  the 
unit  cube.  The  extension  of  the  left  prism  of  Fig.  83 
produces  a  new  prism  shown  in  the  right  of  Fig.  83.  The 
dimensions  of  the  right-hand  figure  are  eXfXb,  and  there- 
fore it  has  /  times  the  volume  of  the  left-hand  figure,  and 
therefore  eXfXb  times  the  volume  of  the  unit  cube.  But 
eXfXb  is  the  product  of  its  three  dimensions. 

Observation.  The  numeric  value  of  the  volume  of  a  rectan- 
gular prism  is  the  product  of  the  numeric  values  of  its  three 
dimensions.  Volume  =  length  X  breadth  X  thickness  =  height  X 
width  X  depth  =  area  of  a  base  X  altitude. 

25.  Volume  of  a  Cube.  A  cube  is  also  a  rectangular 
prism  in  which  the  faces  are  equal  squares  and  the  edges 
are  of  equal  dimensions.  The  volume  (V)  of  a  cube  equals 
the  cube  of  the  length  of  its  edge  (e).  Formulate  this  law 
and  solve  for  e  and  interpret  the  resulting  equation. 

Fig.  84  represents  a  rectangular  prism  ABCDEFGH 
which  has  been  cut  through  the  opposite  edges,  HD  and 


Fig.  84. — Bisection  of  a  Cube. 


FB,  dividing  the  upper  and  lower  bases  into  two  equal 
areas,  by  lines  HF  and  DB  respectively.  The  square 
prism  ABCDEFGH  is  thus  divided  into  two  equal  triangular 
prisms  BCDHFG  and  ABDHEF.  The  triangular  prism 
BCDHFG  is  reproduced  to  the  right  of  the  square  prism, 
in  order  that  it  may  be  shown  that  both  the  square  and  the 
triangular  prisms  have  equal  dimensions.  The  volume 
of  the  triangular  prism  BCDHFG  equals  one-half  the 
volume    of    the    square    prism    ABCDEFGH.     Formulate 


FOEMULAS  OF  MENSURATION 


221 


the  law  for  the  volume  of  a  triangular  prism  in  terms  of 
its  three  dimensions  and  also  in  terms  of  the  area  of  its 
base  and  altitude. 

The  diagonal  of  a  square  prism  is  represented  by  c  in 
Fig.  84.  It  is  also  the  diagonal  of  the  parallelogram 
DBFH.     Show  that  c  is  expressed  by  (27). 


(27) 


c  =  Ve2+62+/2. 


Interpret  (27)  and  solve  for  e,  b,  and  /. 

26.  Volume  of  an  Oblique  Prism.  An  oblique  prism 
differs  from  the  rectangular  prism  in  so  far  as  the  lateral  faces 
are  parallelograms,  some  of  which  are  oblique  to  the  bases. 
In  Fig.  85,  E1F1G1H1ABCD  is  an  oblique  square  prism  stand- 


Hi H 


Gj ,g 


Fig.  85. — An  Oblique  Prism. 


ing  on  the  same  base  A  BCD  as  the  rectangular  square  prism 
EFGHABCD.  It  may  be  imagined  that  the  upper  base 
has  been  slid  along  in  its  own  plane  so  that  FG  takes  the  new 
position  F\G\  and  EH  takes  its  new  position  E\H\.  The 
altitude  of  both  prisms  is  the  same,  as  the  altitude  is  the 
perpendicular  distance  between  the  bases.  Both  prisms 
have  the  same  dimensions.  Both  the  square  and  oblique 
prisms  are  composed  of  one  of  the  two  equal  triangular 
prisms  FiFBCGiG  and  EiEADHiH,  and  both  have  in  com- 
mon the  solid  figure  ABCDHEFiGi.     Therefore  by  sum  of 


222 


PRACTICAL  MATHEMATICS 


parts  axiom,  addition  axiom  and  equality  axiom  we  conclude 
that  the  volumes  of  the  two  square  prisms  are  equal. 

A  prism  has  an  upper  and  a  lower  base  which  are  parallel 
polygons  of  equal  area  and  similar  in  shape.  The  sides 
or  lateral  faces  of  all  prisms  are  parallelograms.  The 
different  prisms  take  their  descriptive  name  from  the  shape 
of  their  bases.  A  right,  i.e.,  a  rectangular  prism  is  one 
having  its  lateral  edges  perpendicular  to  the  bases. 


Fig.  86. — Quadrangular  Prisms. 


27.  The  quadrangular  prisms  in  Fig.  86,  have  an  equal 
altitude  h  and  equal  areas  for  their  bases.  The  two  prisms 
are  divided  into  two  sets  of  equal  triangular  prisms  by 
sections  passed  through  the  diagonals  of  their  bases.     The 


FORMULAS  OF  MENSURATION 


223 


triangular  prisms  of  both  sets  are  respectively  equal.  Right 
and  oblique  triangular  prisms  of  equal  altitude  and  equal 
bases  have  equal  volumes.  Make  a  drawing  to  show  the 
two  triangular  prisms  which  constitute  the  quadrangular 
prism  NOQPJKML. 

28.  Fig.  87  illustrates  a  right  and  an  oblique  triangular 
prism.  They  have  an  equal  altitude  h  which  is  the  perpen- 
dicular  distance   between   the   planes   of   their   respective 


Fig.  87. — Triangular  Prisms. 


bases.  Formulate  the  law  for  the  ratio  of  the  volumes 
of  two  triangular  prisms  of  equal  altitude  in  terms  of  their 
basal  areas  and  also  in  terms  of  their  basal  dimensions. 

29.  Fig.  88  shows  two  pentagonal  prisms  with  equal  and 
similar  bases  and  equal  altitudes.  The  bases  are  divisible 
into  three  mutually  equal  triangles.  The  right  prism  is 
divided  into  three  triangular  prisms  shown  in  the  lower 
view.  The  sum  of  the  volumes  of  the  triangular  prisms 
equals  the  volume  of  the  pentagonal  prism.  The  area  of 
each  triangular  prism  is  the  product  of  its  base  times  altitude. 
Therefore  the  sum  of  the  areas  of  the  triangular  bases 
times  their  common  altitude  equals  the  volume  of  the  pentag- 


224 


PRACTICAL  MATHEMATICS 


onal  prism.  Substitute  for  the  sum  of  the  areas  of  the  bases 
of  the  triangular  prisms  the  area  of  the  base  of  the  pentag- 
onal prisms.  Put  this  proof  in  algebraic  form  designating 
the  area  by  AP,  A\,  A2,  A3  for  the  respective  pentagonal 
and  triangular  prism.     Make  a  drawing  of  the  triangular 


Fig.  88. — Pentagonal  Prisms. 


prisms   which   are   obtained  by   decomposing  the   oblique 
prism  in  Fig.  88. 

30.  Fig.  89  represents  right  and  oblique  hexagonal 
prisms,  with  the  component  triangular  prisms  of  the 
right  prism  in  the  lower  view.  The  two  hexagonal 
prisms  have  an  equal  volume  because  they  have  an  equal 


FORMULAS  OF  MENSURATION 


225 


altitude  and  are  decomposable  into  triangular  prisms  of 
equal  volume.  The  triangular  prisms  may  be  further 
decomposed  into  other  prisms  of  the  same  height,  and 
these  may  be  united  to  form  a  new  total  prism,  differing 
in  shape  from  the  original  figure.     Therefore  two  prisms 


Fig.  89. — Hexagonal  Prisms. 

have  an  equal  volume  when  they  have  equal  altitudes  with 
bases  of  equal  areas.  The  bases  need  not  be  similar  figures. 
For  prisms  of  equal  altitude  the  sum  of  two  or  more  trian- 
gular prisms,  is  to  the  total  prism  or  to  any  other  sum  of  trian- 
gular prisms,  as  the  sums  of  the  respective  areas  of  the  bases. 
Build  new  hexagonal  prisms  from  the  triangular  prisms. 


226 


PRACTICAL  MATHEMATICS 


If  the  bases  of  the  total  prisms  are  regular  polygons 
of  n  sides  then  sections  which  are  constructed  through  the 
axis  and  lateral  edges  will  divide  the  total  prism  into  n  equal 
triangular  prisms.  If  a  and  p  represent  the  apothem  and 
perimeter  of  the  base  of  a  regular  prism  then  its  volume  is 
given  in  (28). 


(28) 


Volume 


aph 


area  of  base  X  altitude, 


31.  Fig.  90  represents  a  right  and  an  oblique  cylinder. 
The  lateral  surface  of  a  cylinder  spreads  into  a  parallelo- 


D 

A 

(        J         r 

M( 

K             \p 

V                •/ 

C               i 

i                   V 

B 

i 

i 

N                 \ 

H 

E 

1 

\r^ 

]T 

V          °           Jg 

r\ 

^^E S 

^-^s^^ 

Fig.  90. — Right  and  Oblique  Cylinders. 


gram  and  its  two  bases  are  equal  parallel  figures.  There- 
fore a  cylinder  may  come  under  the  classification  of  prisms, 
and  accordingly  the  volume  of  a  cylinder  equals  its  base 
times  its  altitude.  Suppose  regular  prisms  were  inscribed  in 
these  cylinders  and  the  number  of  sides  of  the  bases  con- 
tinually increased.  Then  the  area  of  the  polygons  Av, 
representing  the  bases  of  the  prism,  would  gradually  approach 
the  limiting  (lim)  value  which  is  the  area  Ac  of  the  bases 
of  the  cylinder.  The  volumes  Vp,  of  the  inscribed  prisms 
would  gradually  approach  their  limiting  value  which  is  the 
volume  Vc  of  the  cylinder. 


FORMULAS  OF  MENSURATION  227 

v       c  I  construction  of  the 


(30)  lim  AP  =  AC  J  inscribed  prisms. 

(31)  but  Vp  =  hAp  volume  of  a  prism. 

Vp  and  hAp  are  variables,  i.e.,  quantities  continuously 
changing  and  approaching  some  assigned  limit.  Two 
variable  quantities  which  are  always  equal  approach  equal 
limits.  This  is  known  as  the  Fundamental  Theorem  on 
Limits. 

(32)     lim  Vp  —  h  lim  Av    theorem  on  limits. 

/.      (33)     Vc  =  hAc  substitution  (29) (30)  in  (32). 

Observation.  The  interpretation  of  (33)  states  that  the 
volume  of  any  cylinder,  like  the  volume  of  any  prism,  is  the 
product  of  the  area  of  its  base  times  its  altitude. 

Write  a  formula  for  the  volume  of  a  cylinder  in  terms 
of  its  altitude,  perimeter  and  the  radius  of  its  base. 

A  cylinder  is  any  prismatic  figure,  in  which  the  perim- 
eter of  the  base  is  a  closed  curve.  An  element  is  a  line 
in  the  lateral  surface  which  joins  corresponding  points  in 
the  two  bases. 

32.  The  total  surface  of  any  prismatic  figure  equals 
twice  the  area  of  the  base  plus  the  lateral  area.  Formulate 
this  law  and  then  substitute  the  value  of  the  basal  and 
lateral  areas  in  terms  of  the  perimeter  of  the  base.  In  the 
case  of  the  cylinder  what  is  the  total  area  expressed  in  terms 
of  the  radius  of  the  base  and  the  altitude? 

33.  The  Volume  of  a  Pyramid.  A  pyramid  has  one 
base  and  its  lateral  edges  converge  to  a  point  called  the  apex. 

Fig.  91  represents  a  cube  with  its  four  diagonals  17, 
28,  35  and  46  intersecting  in  a  common  point.  The  point 
of  intersection  is  the  common  apex  of  six  equal  pyramids. 
Each  pyramid  has  one  of  the  faces  of  the  cube  for  its  base 


228 


PEACTICAL  MATHEMATICS 


and  its  altitude  is  one-half  the  length  of  an  edge  of  the  cube. 
The  volume  of  the  pyramid  is  one-sixth  of  the  volume  of 
the  cube  of  twice  the  altitude.  Therefore  the  volume  of  a 
pyramid  is  one-third  of  the  volume  of  a  prism  built  on  the 
same  base  and  having  an  equal  altitude.    Therefore  the 


Fig.  91. — Decomposition  of  a  Cube. 

volume  of  a  pyramid  is  expressed  in  (34)  where  h  is  the  alti- 
tude and  A  the  area  of  the  base. 

/0.x     Tr  ,           »             . .     hA     altitude  X  basal  area 
(34)     Volume  of  pyramid  =  —  = = . 

o  o 

Pyramids  are  equal  when  they  have  equal  bases  and  equal 
altitudes. 

34.  Fig.  92  represents  a  cube  12345678  which  has  been 
bisected,  i.e.,  divided  into  two  triangular  prisms  by  the  cut- 


Fig.  92. — Decomposition  of  Prism. 


ting  plane  4268.  In  the  triangular  prism  423867  the  diag- 
onals 82,  27  and  38  have  been  drawn  through  the  three 
lateral  faces  4268,   6237  and  4378  respectively.    Cutting 


FORMULAS  OF  MENSURATION 


229 


planes  passed  through  each  pair  of  diagonals  will  divide 
the  triangular  prism  into  three  equal  pyramids  423-8, 
867-2  and  873-2  as  shown  separately  in  Fig.  93.  Pyramids 
423-8  and  867-2  have  the  equal  bases  423  and  867  and  the 
equal  altitudes  48  and  62.  Pyramid  423-8  may  be  desig- 
nated 438-2.  Pyramids  438-2  and  873-2  have  the  equal 
bases  438  and  837  and  the  equal  altitudes  23.    Therefore 


Fig.  93. — Equivalent  Pyramids. 

the  three  pyramids  are  equivalent  and  the  volume  of  each  is 
equal  to  one-third  of  the  volume  of  the  triangular  prism. 

Observation.  The  volume  of  a  pyramid  is  unaffected  by 
change  in  the  shape  of  the  base  and  the  position  of  its  apex 
provided  the  product  of  its  basal  area  times  its  altitude  remains 
constant. 

35.  Volume  of  a  Cone.     A  cone  may  be  regarded  as  a 

pyramid  with  a  curved  figure  for  its  base.     The  cone  has 

one-third  the  volume  of  a  cylinder  which  has  an  equal 

base  and  altitude. 

/€%^         Tr  .           »             Ah    basal  area  X  altitude 
(35)         Volume  of  cone  =  -5-  = . 


A  line  joining  any  point  in  the  perimeter  of  the  base 
with  the  apex  is  called  an  element. 

Relations  of  Prismatic  and  Pyramidal  Pairs.  Two 
prisms,  two  cylinders,  two  pyramids,  two  cones,  a  prism 
and  a  cylinder  or  a  pyramid  and  a  cone,  have  their  volumes 
in  the  same  ratio  (36)  as  the  products  of  their  respective 
bases  and  altitudes. 


230  PEACTICAL  MATHEMATICS 


™  g-f£. 

(37) 

Vi  A! 
V2    A2 

™  ft-fc 

(39) 

hi_A2 
h2    A\ 

Show  that  for  the  above  pairs  of  figures,  (37)  is  true 
when  they  have  equal  altitudes;  (38)  is  true  when  they  have 
equal  bases;   (39)  is  true  when  they  have  equal  volumes. 

36.  The  total  surface  of  a  pyramidal  figure  equals  the 
area  of  its  base  plus  the  lateral  area.  Express  the  area  of 
a  regular  pyramid  in  terms  of  the  perimeter  and  apothem 
of  the  base  and  in  terms  of  the  slant  height.  Derive  another 
expression  in  which  the  slant  height  is  replaced  by  its  equiv- 
alent in  terms  of  the  altitude  of  the  pyramid  and  the  apothem 
of  the  base.  Express  the  total  area  of  a  right  circular  cone 
in  terms  of  the  radius  of  the  base  and  the  altitude  of  the 
cone.  A  right  pyramid  and  a  right  cone  have  the  position 
of  the  apex  vertically  above  or  below  the  center  of  the  base. 
Any  other  position  of  the  apex  gives  an  oblique  pyramidal 
figure.  A  circular  cone  is  a  pyramidal  figure  with  a  circular 
base. 

37.  Plane  Sections.  In  plane  geometry  lines  may  be 
either  parallel  or  non-parallel.  In  geometry  of  space  a 
third  line  (gooch)  may  be  represented  fulfilling  neither  of 
these  familiar  relations  of  the  plane. 

Three  points  or  their  equivalents  a  point  and  a  line  or 
two  interesting  lines  or  two  parallel  lines  determine  a  plane. 
Two  parallel  planes  have  no  point  in  common.  Two  non- 
parallel  planes  have  a  line  of  intersection. 

Parallel  planes  cut  a  prism,  or  a  cylinder,  or  a  pyramid, 
or  a  cone,  or  a  sphere  in  sections  which  are  similar  figures. 

The  parallel  sections  of  cylindrical  figures  are  equal. 

Pyramidal  figures  having  equal  altitudes  and  equivalent 
bases  have  equivalent  sections  at  equal  distances  from  their 
apexes.  Figures  are  equivalent  when  they  have  equal  areas 
or  volumes. 


FOEMULAS  OF  MENSURATION 


231 


38.  Frustum  of  a  Pyramid    or  Cone.    The  frustum  of 

a  pyramid  or  a  cone  is  the  figure  intercepted  between  the 
base  and  a  parallel  section.  The  volume  is  expressed  in 
(40)  in  which  Ai  and  Au  are  the  lower  and  upper  bases 
respectively 


(40) 


V-$.A,+A,+VAA.) 


For  the  case  of  the  frustum  of  a  right-circular  cone 
show  that  this  simplifies  to 


(41) 


V^jirf+ruZ+nru). 


In  Fig.  94,  I  represents    the   frustum   of    a   triangular 
pyramid,  II  being  the  part  which  has  been  removed.     The 


h'  F' 

Fig.  94. — Frustums. 

volume  of  the  frustum  is  the  difference  between  the  volume 
of  pyramid  A-EFG  and  the  volume  of  pyramid  A-BCD. 
The  base  EFG  is  designated  by  A\  and  the  base  BCD  by 
Au,  and  the  altitude,  i.e.,  the  perpendicular  distance  between 
the  bases  is  designated  by  h.  The  conic  frustum  III  and 
the  pentagonal  frustum  IV  in  Fig.  94  are  equivalent.  Their 
altitudes  and  their  respective  bases  are  equal  in  areas. 

Ex.  5.    Determine  the  formula  for  the  total  surface  of  the 
frustum  of  a  regular  pyramid  in  terms  of  its   basal  areas,  basal 


232  PKACTICAL  MATHEMATICS 

perimeters,  and  the  slant  height.    Write  the  corresponding  formula 
for  the  total  area  of  the  frustum  of  a  right  circular  cone. 

39.  Volume  of  a  Prismoid.  A  prismoid  is  a  figure  in 
which  any  two  parallel  plane  figures  are  joined  by  quadri- 
laterals and  triangles.  The  volume  is  expressed  in  (42)  in 
which  Ai,  Au  and  Am  represent  the  lower,  upper  and  mid 
sectional  areas  respectively,  h  represents  the  altitude  of 
the  figure.     The  majority  of  prismoids  have  quadrangular 


(42)  Volume  of  prismoid  =  -  ( At + Au +4  Am) . 

Deduce  the  formulas  for  the  volume  of  a  prism,  a 
cylinder,  a  pyramid,  and  a  cone  from  (40). 

40.  Volume  of  a  Wedge.  A  wedge  may  be  considered 
as  a  solid  figure  with  a  trapezoidal  base. 

The  usual  upper  base  is  replaced  by  a  line  parallel  to  the 
base.  Each  extremity  of  the  line  is  joined  to  two  adjacent 
vertices  of  the  base  by  the  lateral  edges.  Therefore  a 
wedge  has  five  faces,  two  of  which  are  triangular  and  three 
of  which  are  quadrangular. 

Deduce  the  volume  of  the  wedge  from  the  prismoid 
formula.  Derive  the  volume  of  a  wedge  by  dividing  it  into 
a  triangular  prism  and  a  triangular  pyramid. 

41.  Ungula  of  a  Solid.  A  figure  is  said  to  be  truncated 
when  it  is  cut  by  a  section  which  is  not  parallel  to  the  base 
of  the  figure.  The  portion  of  the  figure  between  the  base 
and  non-parallel  section  is  called  an  ungula.  A  section 
of  a  prism  or  cylinder  which  is  inclined  at  an  angle  0  with 
the  base,  has  an  area  equal  to  the  product  of  the  base  times 
the  secant  of  the  angle. 

The  volume  of  a  cylindric  ungula  shown  in  view  II  of  Fig. 
14  is  equal  to  the  product  of  the  area  of  the  base  times 
the  perpendicular  distance  between  the  base  and  the  center 
of  gravity  of  the  non-parallel  section. 


FORMULAS  OF  MENSURATION  233 

A  cylindric  ungula  is  represented  by  A\FJ  in  Fig.  99. 
The  volume  of  the  cylindric  ungula  which  is  above  the 
section  A1A2  is  given  in  (41).  A\F  =  l  the  length  of  the 
ungula;  FJ  =  b  the  height  of  the  base;  c  =  the  chord  of  the 
base;  D  =  the  diameter  of  the  cylinder;  and  A=the  area 
of  the  segment  FJ,  which  corresponds  to  the  base  of  the 
ungula.  The  base  of  the  ungula  is  a  segment  of  the  circular 
end  of  the  cylinder. 

(43)  Volume  of  ungula=^|  ^-A(D-2b) 

42.  Cylindric  Surface.  A  straight  line  (named  the 
element)  when  moved  parallel  to  itself,  so  that  one  of  its 
extremities  follows  the  perimeter  of  a  plane  figure  generates 
a  cylindric  surface. 

The  development  or  plane  spread  of  a  cylindric  surface 
is  a  rectangle. 

p  =  perimeter  of  the  base  of  a  cylinder  or  prism, 
H  =  length  of  the  element. 

(44)  A  =p#  =  area  of  cylindric  surface. 

Interpret  (44). 

43.  Surface  of  a  Cone.  A  straight  line  (named  the  ele- 
ment) when  moved  so  as  to  pass  through  a  fixed  point,  and 
have  one  of  its  extremities  follow  the  perimenter  of  a  plane 
figure  generates  a  conic  surface. 

The  development  of  a  conic  surface  is  a  circular  sector. 
The  development  of  a  pyramidal  surface  is  a  sector  of  a 
polygon. 

p  —  perimeter  of  the  base  of  a  cone  or  pyramid, 

s  =  the  length  of  the  element, 

(45)  A  =^r  =  area  of  the  surface  of  a  cone. 
Interpret  45. 


234  PEACTICAL  MATHEMATICS 

44.  Figures  of  Revolution.  Any  plane  figure  may  be 
rotated  about  any  line  in  its  plane  and  thereby  generate 
a  figure  of  revolution. 

Sections  of  figures  of  revolution  perpendicular  to  the 
axis  of  rotation  are  either  circles  or  figures  formed  by  con- 
centric circles.     This  principle  is  applied  to  the  lathe. 

A  rectangle  whose  side  is  an  axis  of  revolution  generates 
a  cylinder. 

A  right  triangle  whose  side  is  an  axis  of  revolution 
generates  a  cone. 

A  circle  whose  diameter  is  an  axis  of  revolution  generates 
a  sphere. 

A  parabola  whose  diameter  is  an  axis  of  revolution  gen- 
erates a  paraboloid. 

An  ellipse  whose  diameter  is  an  axis  of  revolution  gen- 
erates a  spheroid  or  an  ellipsoid. 

An  hyperbola  whose  diameter  is  an  axis  of  revolution 
generates  an  hyperboloid. 

Show  that  there  are  two  varieties  for  each  of  the  above 
figures  of  revolution. 

Two  such  rotations  may  be  compounded  and  this  prin- 
ciple may  be  used  upon  the  lathe  to  produce  gun  stocks, 
lasts,  etc. 

That  portion  of  the  perimeter  of  the  rotating  figure 
which  is  not  perpendicular  to  the  axis  generates  a  surface 
of  revolution.  The  line  generates  the  surface  of  revolution. 
The  surface  generates  the  solid  of  revolution. 

45.  Guldinus,  Theorems.  (I)  The  area  of  a  surface 
traced  out  by  the  revolution  of  a  curve  about  an  axis 
in  its  plane,  is  equal  to  the  product  of  the  perimeter  of 
the  curve  times  the  distance  moved  through  by  its  center 
of  gravity. 

(II)  The  volume  generated  by  the  revolution  of  such 
a  curve  is  the  product  of  the  area  enclosed  by  the  curve 
times  the  distance  moved  through  by  the  center  of  area  or 
center  of  gravity. 


FORMULAS  OF  MENSURATION 


235 


46.  Solid  Ring.  Circular  Section.  A  circular  disc 
centered  at  C,  Fig.  95,  rotates  about  the  axis  MN  and 
generates  a  solid  circular  ring. 


Fig.  95. — Section  of  a  Circular  Ring. 

A=the    area  of    the    surface   of   the   ring    and    V  the 

volume  of  the  ring. 
r  =  radius  of  cross-section. 
#  =  the  mean  radius  of  rotation. 


(46) 

A=2%rX2%R, 

(48) 

V  =  rr2X2i:R) 

(47) 

A=±7?rR, 

(49) 

7  =  2x2r2#. 

Interpret  (47)  and  (49). 

47.  Solid  Ring.  Rectangular  Section.  A  rectangle 
centered  at  C,  Fig.  96,  rotates  about  the  axis  MN  and 
generates  a  solid  rectangular  ring. 


k 


Fig.  96. — Section  of  Rectangular  Ring. 

Determine  the  area  and  volume  of  a  solid  ring  of  rectan- 
gular section. 


236 


PRACTICAL  MATHEMATICS 


Ex.  6.    Sphere.    Determine  the  surface  and  volume  of  a  sphere 

by  Guldinus'  Theorems.     The  distance  of  the  center  of  gravity 

4r 
of  a  semicircle  measured  from  the  axis  equals  — . 

ox 


48. 


Fig.  97  represents  a  cylinder,  a  cone,,  and  a  sphere, 
with  equal  dimensions,  i.e.,  the 
height  of  the  cylinder  equals 
the  height  of  the  cone  and  also 
equals  the  diameter  of  the 
sphere.  The  diameter  of  the 
base  of  the  cylinder  and  cone 
equals  the  diameter  of  the 
sphere.  The  cylinder  is  then 
said  to  circumscribe  the  cone 
and  the  sphere.  Designating 
the  volumes  of  a  cylinder,  a 
sphere  and  a  cone  by  Vcv, 
Vs,  VCo  respectively,  show  that 
(50)  expresses  their  relative 
magnitudes  when  the  cylin- 
der circumscribes  the  cone  and 
sphere. 


Fig.  97. — Volume  Relations. 


(50) 


Vev  :  Vs:  Vco  =3  :2  :1 


49.  When  a  cylinder  or  cone  is  represented  in  perspective 
as  shown  in  Fig.  97,  the  bases  are  represented  as  ellipses.  The 
construction  of  the  ellipse  is  shown  at  the  upper  base  of  the 
figure.  The  center  A  of  the  upper  base  is  also  the  center 
of  a  minor  circle  CMLQ,  whose  radius  is  the  semiminor 
axis  of  the  desired  ellipse,  and  A  is  also  the  center  of  a  major 
circle  BVTD  whose  radius  is  the  semimajor  axis  of  the  de- 
sired ellipse. 

From  A  draw  radial  lines  AV  and  AT,  intersecting  the 
minor  circle  at  M  and  L  respectively,  and  the  major  circle 
at  V  and  T  respectively.     The  intersection  Af  of  a  horizontal 


FORMULAS  OF  MENSURATION  237 

line  through  M  with  a  vertical  line  through  V  is  a  point  on 
the  ellipse.  The  intersection  P  of  a  horizontal  line  through 
L  with  a  vertical  line  through  T  is  a  point  on  the  ellipse. 
The  points  C,  N,  P,  D,  determine  a  quadrant  of  the  ellipse. 
The  other  quadrants  are  symmetrical  to  the  major  and 
minor  axes.  The  distance  RS  =  RP,  PS  being  perpendicular 
to  AD. 

Ex.  7.  Show  that  the  surface  of  a  sphere  equals  four  times 
the  area  of  a  section  passing  through  the  center  of  a  sphere. 

50.  Segment  of  a  Sphere.  The  segment  of  a  sphere  is 
the  volume  cut  off  by  any  section.  A  line  drawn  perpendic- 
ular to  the  section,  i.e.,  the  base  of  the  segment  and  extending 
to  the  surface  is  the  altitude  of  the  segment.  Formula 
(51)  expresses  the  volume  of  a  spherical  segment  in  which 
r  is  the  radius  of  its  base.  (52)  expresses  the  area  of  the 
spherical  segment  in  which  R  is  the  radius  of  the  sphere. 

(51)  Volume  of  spherical  segment = ■=  h  (  r2 +—  j . 

(52)  Area  of  spherical  segment  —  2%Rh. 

Interpret  (51)  and  (52).     Solve  (51)  for  r. 

51.  The  volume  of  a  spherical  zone  is  given  in  (53) 
in  which  n  and  r2  are  the  radii  of  the  respective  bases  and 
h  is  the  altitude  of  the  zone. 


(53)        Volume  of  spherical  zone  =  ^  h  (  n2 -{-r2 


•+?>• 


Show  that  (53)  may  be  derived  from  the  formula  of 
Ex.  6,  and  formula  (51).  Interpret  (53)  and  solve  for 
ri  and  V2. 

52.  Oblate  and  Prolate  Spheroids.  When  an  ellipse  is 
rotated  about  its  minor  axis  it  generates  an  oblate  spheroid, 
and  when  it  is  rotated  about  its  major  axis  it  generates  a 
prolate  spheroid.     Derive  the  formulas  for  the  surface  and 


238  PRACTICAL  MATHEMATICS 

volumes  of  these  figures  of  revolution  by  Guldinus'  Theorems. 

4 
The  center  of  gravity  of  a  semicircle  equals  ~  times  the 

ox 

semi-diameter  on  which  it  is  measured  from  the  center. 

53.  The  volume  of  a  segment  of  a  spheroid  whose 
base  is  perpendicular  to  one  of  the  axes,  is  expressed  in  (54) 
when  the  base  is  circular,  and  in  (55)  when  the  base  is  ellip- 
tical. Di  is  the  diameter  of  the  polar  axis,  i.e.,  the  axis 
of  revolution  and  Z>2  is  the  equatorial  diameter,  i.e.,  the  diam- 
eter perpendicular  to  the  axis  of  rotation,  h  represents  the 
altitude  of  the  segment. 

'-(f-iXf)* 

54.  Volume  of  a  Paraboloid.  The  volume  of  a  para- 
boloid equals  one-half  the  volume  of  a  circumscribing 
cylinder.  Formulate  the  law  and  interpret  the  formula. 
Write  fcne  formula  for  the  frustum  of  a  paraboloid. 

55.  Volume  of  an  Hyperboloid.  By  means  of  the 
prismoidal  formula  (42)  show  that  the  volume  of  an  hyper- 
boloid is  expressed  in  (56).  n  is  the  radius  of  the  base; 
T2  is  the  radius  of  the  middle  section;  h  is  the  altitude  of 
the  paraboloid. 

(56)  Volume  of  hyperboloid  =  -(ri2+4r22)/i. 

Interpret  (56)  and  solve  for  n  and  r2. 

56.  Regular  Solids.  A  regular  polyhedron,  i.e.,  regular 
solid  has  its  faces  and  edges  equal  and  its  plane  and  solid 
angles  equal.     There  are  only  five  regular  solids. 

The  tetrahedron  is  a  regular  triangular  pyramid  whose 
sides  are  equilateral  triangles.  It  has  six  edges  and  four 
faces. 


FORMULAS  OF  MENSURATION  239 

The  hexahedron  is  a  cube.  It  has  twelve  edges  and  six 
faces. 

The  octahedron  is  a  solid  which  is  bounded  by  equilateral 
triangles.     It  has  twenty  edges  and  eight  faces. 

The  dodecahedron  is  a  solid  which  is  bounded  by  regular 
pentagons.     There  are  twenty  edges  and  twelve  faces. 

The  icosahedron  is  a  solid  bounded  by  equilateral 
triangles.     There  are  thirty  edges  and  twenty  faces. 

With  the  exception  of  the  tetrahedron  the  other  poly- 
hedrons have  their  opposite  faces  and  edges  in  parallel 
pairs. 

Ex.  8.  Construct  the  development  or  spread  for  the  five  poly- 
hedrons. Determine  the  volume  of  a  tetrahedron  with  a  unit 
edge. 

57.  Volume  of  an  Irregular  Solid.  Any  rule  which 
is  applicable  to  the  determination  of  an  irregular  area  may 
be  applied  to  the  determination  of  a  volume.  A  base  line 
is  drawn  in  the  base  of  the  solid  so  as  to  represent  the  hori- 
zontal projection  of  the  greatest  dimension  of  the  solid. 
The  base  line  is  then  divided  into  equal  spaces  and  at  the 
points  of  division  of  the  base  line  a  series  of  parallel  vertical 
planes  are  extended  through  the  solid.  The  planes  cut  a 
series  of  parallel  sections  of  more  or  less  irregularity  through 
the  solid.  Their  area  is  determined  by  any  of  the  rules 
for  area.  The  mean  of  these  areas  multiplied  by  the  length 
of  the  base  line  gives  the  volume  of  the  solid.  If  the  mag- 
nitudes of  the  measured  areas  are  scaled  and  laid  off  as 
ordinates  at  the  corresponding  points  of  division  on  the 
base  line,  a  new  irregular  figure  is  formed  whose  area  expresses 
the  magnitude  of  the  volume  of  the  solid. 

Ex.  9a.    Area-linear  Error. 

An  error  of  5  lbs.  in  the  weight  of  1  mile  of  Cu  wire  causes 
what  per  cent  error  in  the  calculated  value  of  the  diameter? 

Ex.  9b.    Linear-square  Error. 

The  length  of  a  seconds  pendulum  is  increased  by  10*oo  part. 
How  many  seconds  will  the  clock  lose  in  a  day? 


240  PKACTICAL  MATHEMATICS 

The  time  of  a  complete  oscillation  T  =2%x  — ,  where  I  is  the 

\  9 
length,  and  g  is  constant  for  gravity. 

Ex.  9c.    Linear-volume  Error. 

The  radius  of  a  sphere  is  found  by  measurement  to  be  5  ins. 
What  error  will  be  caused  in  the  calculated  value  of  the  volume 
by  an  error  of  1  per  cent  in  the  measured  value  of  the  radius? 

Ex.  10.     Mixed  Error. 

The  value  M  of  the  magnetic  moment  of  a  magnet  is  calcu- 
lated from  the  two  formulas 

M     (rf«-ft)t 

(6)    h — ; d~' 

Eliminate  H. 

L,  d,  I,  and  T  are  observed  quantities.  An  error  of  2  per 
cent  excess  is  made  in  observing  T  and  all  other  readings  are 
correct  to  0.1  per  cent.  What  is  the  approximate  per  cent  error 
in  the  calculated  value  of  Ml 

58.  The  Binomial  Theorem.  The  binomial  theorem 
gives  the  following  formula  for  expanding  a  binomial  to 
any  power.  An  expansion  is  the  complete  or  partial  state- 
ment of  terms  which  result  when  an  indicated  operation  has 
been  performed. 
(a+br  =  an+m^b+njr^an^+(n)  (n-1)  (n-2) 

on_363+       n(n-l)...(n-K+l)a»ZV+       ^ 

a  and  b  represent  any  two  quantities  and  n  is  any  expo- 
nent which  may  be  integral  or  fractional  and  positive  or 
negative.  In  like  manner  a  and  b  may  have  opposite 
signs. 

Ex.  11.  Expand  (a+b)n  for  each  of  the  following  values  of 
n,  viz.,  1,  2,  3,  4,  5,  6,  7. 

Ex.  12.    Approximations. 

If  we  allow  the  approximation  a2+2ab  for  (a+6)2,  what  is 
the  per  cent  error  (a)  when  a  =6  =  1;    (6)  when  a  =  3,  and  6=2; 


FORMULAS  OF  MENSURATION  241 

(c)  when  a  =  10  and  6  =  1;  (d)  when  a  =  100  and  6  =  .l.  What 
are  the  conditions  for  least  error? 

Ex.  13.  (l+x)(l+y)  =?.  If  we  allow  the  approximation 
1+x+y, 

(a)  What  is  the  per  cent  error  when  x  =  1 ;  y  =  1? 

(6)  What  is  the  per  cent  error  when  X  =0.1;  y  =0.1? 

(c)  What  is  the  per  cent  error  when  x  =  100;  y  =  100? 

What  are  the  conditions  for  least  error? 

Ex.  14.  (1  +x){l  +y)(l  +z)  =?  If  we  allow  the  approximation 
1+x+y+z, 

(a)  What  is  the  per  cent  error  when  x  =  1  =y  =z? 

(6)  What  is  the  per  cent  error  when  x  =  l=y;  z  =  .l? 

(c)  What  is  the  per  cent  error  when  x  =  l;  y  =  .l=z? 

(d)  What  is  the  per  cent  error  when  x  =  l;  y  =  .l;  z  =  .01? 
What  are  the  conditions  for  least  error? 

Ex.  15.     (1  +x)n  =  ?     If  we  allow  the  approximation  1  +nx, 

(a)  What  is  the  per  cent  error  when  x  =  1 ;  n  =  1? 

(b)  What  is  the  per  cent  error  when  x  =  1 ;  n  =  100? 

(c)  What  is  the  per  cent  error  when  x  =  100;  n  =  1? 

(d)  What  is  the  per  cent  error  when  z  =5;  n  =5? 
What  are  the  conditions  for  least  error? 

Ex.  16.     Ship  Area. 

The  semiordinates  of  the  load  water  plane  of  a  vessel  are 
.2,  3.6,  7.4,  10.  11,  10.7,  9.3,  6.5,  2  ft.,  respectively,  and  they  are 
15  ft.  apart.  What  is  the  area?  Compute  by  different  methods 
and  tabulate. 

Ex.  17.    Area  Curve. 

The  ordinates  of  a  curved  figure  in  inches  are  2.6,  3.5,  3.66, 
3.63,  3.37,  2.85,  2.4,  2.1,  1.89,  1.74,  1.6,  1.38,  .49.  The  common 
intervals  =  \  in.    Determine  the  area  below  the  curve. 

Ex.  18.    Area  Indicator  Diagram. 

The  length  of  an  indicator  diagram  is  4  ins.,  the  end  ordinates 
are  1  and  .22,  and  the  other  ordinates  following  the  first  are  1, 
.82,  .71,  .55,  .45,  .38,  .33,  .29,  .26  in.  respectively.  The  scale  of 
pressure  is  60  lbs.  equals  1  in.  Determine  the  mean  pressure  (a) 
by  the  planimeter;  (b)  by  Simpson's  rule;  (c)  by  mid  ordinate 
rule;   (d)  by  counting  squares. 

Ex.  19.  The  half  ordinates  of  the  midship  section  of  a  vessel 
are  12.8,  12.9,  13,  13,  13,  12.9,  12.6,  12,  10.5,  6,  1.5  ft.  respec- 
tively. The  common  distance  between  the  ordinates  is  18  ins. 
What  is  the  area? 

Ex.  20.  The  base  of  a  prism  is  a  triangle  whose  sides  are  27,  25, 
and  38  ft.  respectively  and  whose  height  is  10  ft.   What  is  its  volume? 


242  PRACTICAL  MATHEMATICS 

Ex.  21.    Volume  of  Stream. 

A  section  of  a  stream  is  10  ft.  wide  and  10  ins.  deep.  The 
mean  flow  of  the  water  through  the  section  is  3  miles  an  hour. 
(a)  How  many  gallons  flow  through  the  section  per  hour?  (6) 
per  day? 

Ex.  22.     Solid  vs.  hollow  pillars. 

What  is  the  weight  of  a  hollow  steel  pillar  10  ft.  long,  whose 
external  diameter  is  8  ins.,  and  internal  diameter  4  ins.?  What 
is  the  diameter  of  a  solid  pillar  of  the  same  weight  and  length? 
One  cubic  foot  steel  weighs  490  lbs. 

Ex.  23.    Weight  of  Cable. 

A  single-core  electric  cable  consists  of  a  cylindric  copper  wire 
surrounded  by  a  coating  of  insulation  and  an  outer  coating  of 
lead.  The  area  of  the  cross-section  of  the  copper  is  .25  sq.in. 
The  thickness  of  insulation  is  .11  in.,  and  the  thickness  of  the  outer 
covering  is  .11  in.     What  is  the  diameter  and  weight  of  the  cable? 

1  cu.in.  copper  weighs  .32  lb. 

1  cu.in.  lead  weighs  .41  lb. 

1  cu.in.  insulation  weighs  .034  lb. 

Ex.  24.     Brass  Ball. 

A  hollow  sphere  of  brass  is  found  to  weigh  50  lbs.  Its  external 
diameter  is  10  ins.    What  is  the  internal  diameter? 

1  cu.in.  brass  weighs  .3  lb. 

Ex.  25.    Railway  Embankment. 

A  railway  embankment  is  12  ft.  high.  The  top  is  28  ft.  wide 
and  sides  slope  1  :  2  to  the  horizontal.  How  many  cubic  yards 
of  earth  are  required  to  continue  the  embankment  1000  ft.? 

Ex.  26.  A  bus  of  copper  bar  3£  ins.  by  |  in.  is  35  ft.  in  length. 
What  is  its  resistance?  What  length  of  the  bar  will  serve  as  an 
ammeter  shunt  to  give  45  milli volts  for  a  current  of  2000  amp. 
What  is  the  total  weight  of  the  bus? 

Ex.  27.  Determine  the  number  of  cubic  yards  in  a  bank  of 
earth  on  a  horizontal  rectangular  base  60  ft.  by  20  ft.  The  four 
sides  of  the  bank  slope  to  a  ridge  at  an  angle  of  40°  to  the  horizontal. 

Ex.  28.    Storage  of  Water. 

A  cylindrical  vessel  16  ft.  diameter,  20  ft.  long  is  filled  with 
water.    What  is  the  weight  of  water  in  tons? 

One  cubic  foot  of  water  weighs  62.5  lbs. 

Ex.  29.  A  prism  has  a  base  of  50.32  sq.ins.  What  is  the  area 
of  a  section  inclined  20°  to  the  horizontal? 

Ex.  30.  The  base  of  a  right  cone  is  an  ellipse  whose  axes  are 
21  ft.  and  14  ft.  respectively.  The  altitude  is  12  ft.  What  is 
its  volume? 


FORMULAS  OF  MENSURATION  243 

Ex.  31.  A  right  circular  cone  is  divided  by  two  planes  parallel 
to  the  base.  These  planes  are  equidistant  from  the  base  and  apex. 
Compare  the  three  volumes  of  the  divided  solid.  The  cone  has 
a  radius  of  28  ins.  and  an  altitude  of  30  ins.  The  sections  are 
separated  by  12  ins. 

Ex.  32.     Comparison  of  Surfaces. 

Compare  the  surface  of  a  cube,  cylinder,  sphere,  and  cone,  the 
volume  in  each  case  being  1  cu.ft.  The  altitudes  of  cones  and  cyl- 
inders are  to  be  considered  equal  to  the  diameters  of  their  bases. 

Ex.  33.     Comparison  of  Volumes. 

A  cone,  cylinder,  sphere,  cube,  have  equal  surfaces.  Determine 
their  volumes.  Cones  and  cylinders  are  to  have  equal  altitudes 
and  diameters. 

Ex.  34.  Compare  the  surface  and  volume  of  a  sphere  and  cube 
of  equal  dimensions. 

Ex.  35.     Funnel. 

A  tin  funnel  is  3.5  ins.  diameter  and  has  a  slant  height  of  7 
ins.  How  much  material  is  required  in  the  manufacture  of  its 
conic  surface  allowing  3  per  cent  for  overlapping. 

Ex.  36.  Determine  the  volume  of  an  incandescent  lamp  bulb 
from  its  measurements.  Check  the  result  by  filling  the  bulb 
with  a  measured  and  weighed  powder.  Check  also  by  immersing 
the  bulb  in  a  graduated  beaker  containing  water. 

Ex.  37.     Cylindrical  Boiler. 

Determine  the  volume  of  water  in  a  horizontal  cylindrical 
boiler  of  length  L,  diameter  D,  when  the  height  of  the  water  is 
h.    In  the  answer  express  the  value  of  the  angle  as  an  anti-function. 

Ex.  38.     Governor  Ball. 

A  sphere  of  radius  R  is  pierced  by  a  cylindrical  hole  through 
the  center,  r  is  the  radius  of  the  cylinder.  What  is  the  volume 
of  the  ball? 

Ex.  39.  What  would  be  the  equivalent  circle  for  a  lake  whose 
area  is  3  acres?  How  would  you  determine  the  area  of  the  lake 
from  the  contour  on  a  map?  What  is  the  equivalent  number  of 
sections  (land  measure)?  How  would  you  determine  the  area  and 
volume  of  a  lake  from  a  topographic  map? 

Ex.  40.  A  running  track  is  in  the  form  of  a  link  enclosing  a 
football  field  with  10  ft.  margin  and  has  circular  ends.  Can  such 
a  track  be  constructed  4  laps  to  the  mile? 

Ex.  41.  With  a  radius  of  3  ins.  construct  an  arc  subtending 
a  chord  of  3  ins.  What  is  the  length  of  the  arc?  Determine  the 
length  accurately  and  compare  the  result  with  that  obtained  by 
approximate  methods. 


244  PRACTICAL  MATHEMATICS 

Ex.  42.  A  dome  is  in  the  form  of  a  segment  of  a  sphere  of 
radius  100  ft.  The  height  of  the  dome  is  80  ft.  How  many 
cubic  yards  of  cement  1  in.  thick  will  be  required  to  cover  it? 

Ex.  43.  A  basin  is  in  the  form  of  a  zone  of  a  sphere.  It  is  8  ins. 
deep.  The  bottom  is  a  circle  of  4  ins.  diameter.  The  top  is  a  circle 
of  15  ins.  diameter.     How  much  water  will  it  hold  at  each  inch  level? 

Ex.  44.  What  is  the  thickness  of  lead  pipe  2.5  ins.  internal 
diameter  if  it  weighs  7.862  lbs.  per  linear  foot? 

Ex.  45.  What  is  the  weight  of  an  iron  pipe  12  ft.  long,  14  ins. 
of  external  diameter,  13  ins.  internal  diameter  given  the  weight 
of  1  cu.in.  of  iron  =0.27  lb. 

Ex.  46.  What  is  the  weight  of  a  cast-iron  bar  1  in.  X 1  in.  X 1  yd.? 
What  error  would  be  made  by  calling  this  weight  10  lbs.? 

Ex.  47.  Water  flows  at  the  rate  of  4.96  ft.  per  sec.  through 
a  cylindrical  pipe  11  ins.  in  diameter.  What  is  the  supply  in  gallons 
per  minute?    7.5  gals.  =  1  cu.ft. 

Ex.  48.  The  outside  diameter  of  a  road  roller  is  3  ft.  and  its 
outside  width  4  ft.  The  metal  is  2  ins.  thick  on  the  surface.  It  is 
closed  at  the  ends  which  are  1  in.  thick.  The  axle  and  handle 
weigh  11  lbs.  and  the  metal  of  which  the  roller  is  made  weighs 
437  lbs.  per  cubic  foot.    What  is  the  weight  of  the  outfit? 

Ex.  49.  How  many  bricks  will  be  required  to  build  a  foundation 
4  ft.  X5  ft.  X4  ft.?  Each  brick  measures  9  ins.  X3  ins.  X4.5  ins., 
including  mortar. 

Ex.  50.  A  solid  sphere  has  a  cylindrical  hole  drilled  centrally 
through  it.  If  2b  be  the  length  of  the  bore,  show  that  the  remain- 
ing volume  is  equal  to  that  of  a  sphere  of  radius  b. 

Ex.  51.     Application  of  the  Pythagorean  Theorem. 

On  a  piece  of  cross-section  paper  construct  a  right  triangle 
whose  sides  are  3,  4,  and  hypotenuse  5.  Build  a  square  on  each 
side  and  on  the  hypotenuse  and  count  the  area.  It  will  be  found 
that  the  sum  of  the  areas  on  the  two  sides  equals  the  area  on  the 
hypotenuse.     Try  this  again  for  a  9-12-13-right  triangle. 

But  these  facts  are  also  substantiated  by  the  theorem  of  the 
algebraic  squares.  Construct  a  right  triangle  with  sides  equal 
to  one.    What  is  the  length  of  the  hypotenuse? 

Ex.  52.  Construct  lines  corresponding  to  V'3,  V5,  V7,  V8, 
VlO,  V.  1  and  prove. 

Show  how  these  lines  could  be  constructed  by  using  a  theorem 
on  the  circle. 

Ex.  53.  What  is  the  range  of  area  of  the  earth's  surface  which 
may  be  illuminated  by  a  searchlight  mounted  in  the  top  of  a  550- 
ft.  tower. 


FORMULAS  OF  MENSURATION 


245 


Ex.  54.  A  circle  25  ins.  in  diameter  is  to  be  divided  into  four 
equal  areas  by  concentric  circles.    What  are  their  diameters? 

Ex.  55.  Fig.  98  represents  the  cross-section  of  a  drum  of  a 
boiler.    The  drum  is  40  ins.  in  diameter  and  its  cylindric  portion 


Fig.  98. — Cross-section  of  a  Steam  Drum. 

is  11  ft.  10  ins.  in  length.  The  ends  of  the  drum  are  spherical 
segments  with  a  40-in.  radius  and  extend  7  ins.  beyond  the  end 
of  the  cylindric  portion.  Half  the  contents  of  the  drum  is  evaporated 
in  25  min.  Determine  the  level  of  the  liquid  at  the  end  of  each 
minute  assuming  the  same  rate  of  evaporation  to  continue.  What 
error  will  be  made  if  the  volume  of  the  spherical  ends  is  considered 
negligible? 


Fig.  99. — Cylindric  Tank. 

Ex.  56.    Fig.  99  represents  a  cylindric  vessel  of  length  L  and 
diameter  D.    The  ends  are  spherical  segments  of  radius  R  and 


246 


PRACTICAL  MATHEMATICS 


height  a.  The  vessel  is  tipped  at  an  angle  6  when  measured  to 
a  horizontal  plane.  The  height  of  the  liquid  above  the  plane 
is  designated  by  hi,  h2,  hs,  hi.  Determine  the  volume  of  the 
liquid  for  the  levels  AXA2,  ByB2,  CiC2,  and  EiE2. 

Ex.  57.  Fig.  100  represents  two  views  of  a  commutator  bar. 
Determine  the  volume  and  weight  of  432  bars  like  ABCEGKLN 
allowing  .32  lb.  per  cubic  inch.  The  lower  edge  of  the  bar  ends 
along  the  line  KG.  The  inside  diameter  of  the  commutator  has 
a  diameter  =  1  ft.  10 J  ins.  Determine  the  thickness  of  the  mica 
insulation.    The  cross-section  of  a  bar  is  a  trapezoid.    The  bar  has 


b  r    s 


v  w  x  Y 

Fig.  100. — A  Commutator  Bar. 


1 


T  U 


the  following  dimensions:  AB  =  8M  ins.;  NM  =2.55  ins.;  DC  =3.11 
ins.;  KJ  =2.00  ins.  =HG;  FQ=JH  =3.15  ins.;  FH  =21  ins.; 
TO  =  1.56  ins.;  DE  =  .W  in.;  ##  =  1.15  ins.;  QM  =  1.59  ins.; 
ML  =  .13  in.;  LJ  =  1.15  ins.  The  thickness  at  B  =  .1769  in.;  the 
thickness  at  £  =  .1351  in.;  angle  1  =3=3°;  angle  2=4=30°. 

Ex.  58.  Determine  the  volume  of  288  bars  like  ABCEYVPLN 
in  Fig.  100.  The  inside  diameter  of  the  commutator  is  llxl  ins. 
The  section  of  a  commutator  bar  is  trapezoidal.  Determine  the 
thickness  of  the  mica  insulation.  The  dimensions  of  the  bar  are  as 
follows:  A5  =  10Mins.;  AQ  =1.36  ins.;  QF  =  6.65  ins.;  FB=2.9Q 
ins.;  FX=2£ms.;  FD  =  1.20  ins.;  DE  =  .15m.;  EX  =  1.18  ins.; 
QM  =  1.28  ins.;  ML  =  .07  in.;  LO  =  .78  in.;  OIF  =  .40  in.;  the 
thickness  at  #  =  .1528  in.;  the  thickness  at  F  =  .0976  in.;  angles 
1=3=3°;  angles  2  =4  =30°, 


CHAPTER  XII 
THE    QUADRATIC    EQUATION 

1.  A  foemula  or  equation  may  contain  a  number  of 
letters  or  letters  and  numeric  quantities.  When  the  given 
equation  is  cleared  of  fractions  and  radicals  we  can  determine 
the  degree  of  the  equation  for  each  of  its  elements,  i.e.,  its 
letters.  Each  element  may  be  considered  an  unknown 
quantity,  i.e.,  a  letter  whose  value  is  to  be  expressed  numer- 
ically and  literally  in  terms  of  the  other  elements  of  the 
formula. 

If  an  element  enters  a  formula  with  no  exponent  other 
than  one  then  the  formula  is  a  simple  equation,  while  con- 
sidering that  element  only.  It  is  called  a  first  degree  or 
linear  equation  in  other  parts  of  this  text. 

A  quadratic  formula  is  an  equation  which  contains  the 
second  power  of  an  unknown  quantity  and  is  therefore 
called  a  second  degree  equation.  Such  an  equation  usually 
contains  a  term  with  the  first  power  of  the  unknown  and 
an  absolute  term,  i.e.,  a  term  without  the  unknown  quan- 
tity. Either  of  the  latter  may  be  missing  from  a  quadratic 
equation. 

abu+agv 

Eq.  (1)  is  a  quadratic  equation  for  each  of  the  elements, 
u  and  v  but  a  simple  equation  for  each  of  the  elements 
t,  a,  b,  and  g. 

247 


248  PEACTICAL  MATHEMATICS 

Quadratic  equations  are  solved  by  factoring  when  the 
formula  contains  one  element  only.  A  method  known  as 
completing  the  trinomial  square  is  resorted  to  in  other  cases 
but  this  method  may  be  abbreviated  by  the  use  of  a  working 
formula.  Graphic  methods  may  also  be  used  for  the  solu- 
tion of  a  quadratic  equation. 

2.  Solving  a  Quadratic  Equation  by  Factoring.  Multiply 
the  coefficient  of  the  second  power  of  the  unknown  by  the 
absolute  term.  Resolve  this  product  into  two  factors, 
whose  sum  equals  the  coefficient  of  the  first  power  of  the 
unknown.  Replace  the  latter  by  its  equal  and  factor  by 
grouping. 

(1)  5p2+18p+16  =  0; 

5X16  =  80,     80  =  10X8,     10+8  =  18,     18p  =  10p+8p 

(2)  5p2+10p+8p+16  =  0  subs,  for  18p  in  (1) 

(3)  hp  {p + 2)  +  8  (p + 2)  =  0  factoring  in  groups 

(4)  (5p+8)  (p+2)  =  0  factoring  H.C.F. 

(5)  (5p+8)=0  div.  (4)  by  (p+2) 

g 

(6)  p  =  —  £-        trans,  and  div.  in  (5) 

o 

(7)  p+2  =  0  div.  (4)  by  (5p+8) 

(8)  p=-2         trans,  in  (7) 

Steps  (5)  and  (7)  could  be  omitted  if  the  solutions  of 
p  expressed  in  (6)  and  (8)  are  written  by  inspection  of  (4). 
The  two  values  of  p  in  (6)  and  (8)  are  written  from  the 
factors  of  (4)  by  changing  the  signs  of  the  numeric  terms 
and  dividing  the  latter  by  the  corresponding  coefficient  of  p. 


THE  QUADRATIC  EQUATION  249 

Solve  the  following  examples  by  factoring: 

Ex.  1.    5p2+24p+16=0. 

Ex.  2.     2a2+4a  +  16=0. 

Ex.  3.    6z2-13z-63=0. 

Ex.  4.    c2+3.7c+3=0. 

Ex.  5.     12fc2-26&+10=0. 

As  a  check  on  the  work  we  should  substitute  the  value  of  the 
unknown  in  the  original  equation. 

3.  Solving  a  Quadratic  Equation  by  Completing  the 
Square.  This  method  may  be  summarized  by  five  distinct 
steps : 

(I)  Transpose  the  first  and  second  powers  of  the  unknowns 
to  the  left  member  and  the  knowns  to  the  right  member  and 
collect. 

(II)  Divide  the  equation  by  the  coefficient  of  the  second 
power  of  the  unknown. 

(III)  Complete  the  trinomial  square  in  the  left  member 
by  adding  to  the  equation  the  square  of  half  the  coefficient 
of  the  first  power  of  the  unknown. 

(IV)  Perform  the  square  root  upon  the  equation,  indicat- 
ing the  root  in  the  right  member  with  the  ambiguous  sign 
(=b)  which  is  read  plus  or  minus. 

(V)  Transpose  and  simplify  the  solution. 

(2)  5p2+24p+16  =  0  The   given  equa- 

tion 

(I)  5p2+24p  =  - 16         Trans,  in  (2) 

(ID  p2+fp=-y         Div.  (I)by5 

/TTTN      2.24       .    f12Y       144        16       64  A^/12V.       /TTX 

(III)  P2+TP+(T)  =  25—  y  =  25  AddU)    t0  (II) 


250 


PRACTICAL  MATHEMATICS 


(IV) 


(V) 


(VI) 


p+ 


12 


/64 
:\25 


4 


12  ,  8 

-12+8 


12±8 


P  = 


and  p  = 


5 
12 


5 

_4 

5 

-4 


Root  of  (III) 


Trans.  (IV) 


simplifying  (V) 


Ex.  6.  Check  the  results  of  Ex.  1,  2,  3,  4,  and  5  by  the 
method  of  completing  the  trinomial  square. 

Observation.  If  the  value  of  the  unknown  is  substituted, 
it  should  reduce  the  original  equation  to  zero  when  simplified. 
Those  values  of  the  unknown  which  reduce  an  equation  to 
zero  form  its  factors  when  connected  to  the  unknown  by  a  minus 
sign. 

4.  Every  equation  of  the  second  degree  may  be  con- 
sidered as  a  special  form  of  (3)  by  substituting  definite 
literal  or  numeric  values  for  a,  b  and  c,  and  the  correspond- 
ing unknown  to  replace  x. 


(3) 


ax2+bx+c  =  0. 


Any  equation  derived  from  (3)  will  also  validate  the  like 
values  substituted  for  a,  b  and  c.  Therefore,  solving  (3) 
by  the  method  of  completing  the  square,  we  obtain  (7) 
which  is  a  working  formula  for  all  quadratic  equations. 


(4) 
(5) 
(6) 
(7) 


x2-\ — X  = 
a 


X^aX^\2a)      4a2    a 


c     b2  —  £ac 


x+ 


2a 


± 


4a2 
V62-4ac 
2a       ' 


-b  .  V62-4ac     -6=bV62-4ac 


2a 


2a 


2a 


THE  QUADRATIC  EQUATION  251 

Given  any  equation,  whose  coefficient  of  the  (unknown 
squared)  is  a,  coefficient  of  (unknown)  is  b,  and  absolute 
term  is  c,  then  the  unknown  is  given  in  the  formula  (7)  in 
terms  of  the  same  constants  a,  b,  and  c.  The  formula 
should  not  be  used  until  the  student  is  thoroughly  familiar 
with  its  derivation  so  that  he  may  apply  it  skillfully. 

5.  When  the  roots,  i.e.,  the  values  of  the  unknown  of  an 
equation  are  given,  and  we  wish  to  construct  the  quadratic 
equation,  we  may  use  the  principle  stated  in  the  observation 
under  (3).     Suppose  the  roots  are  m  and  n,  then 

(8)  (x-m)  =  0  and  (x-n)  =  0; 

(9)  (x-m)  (x-n)=0; 

(10)  x2—(m+ri)x+mn  =  0; 
but 

(4)  *2+hi+Ca=°- 

Comparing  (10)  with  (4)  we  observe, 

(11)  m+n=--  (12)     mn  =  -. 

a  a 

The  sum  of  the  roots  equals  the  negative  ratio  of  the 
coefficients  b  to  a. 

The  product  of  the  roots  equals  the  ratio  of  the  absolute 
term  c  to  a. 

6.  The  Graphic  Construction  of  the  Roots  of  a  Quadratic 
Equation.  The  roots  of  a  quadratic  equation  may  be  con- 
structed graphically  upon  a  sheet  of  cross-section  paper. 
Lay  off  AB  =  c;  on  a  perpendicular  to  AB  from  B  lay 
off  BC  =  b;  on  a  parallel  to  AB  from  C  lay  off  CD  =  a.  These 
lines  are  laid  off  cyclically  in  a  counter-clockwise  direction 
and  have  the  same  sense  (positive)  as  shown  in  the  circle. 


252 


PRACTICAL  MATHEMATICS 


A  similar  circle  for  counter-clockwise  or  negative  sense  will 
aid  in  laying  off  negative  values  for  a,  b  and  c. 


Fig.  101. — Graphic  Construction  of  the  Roots  of  a  Quadratic  Equation. 

Join  D  with  A,  and  bisect  DA  at  0;  with  0  as  center 
anpl  OD  =  OA  =  radius,  draw  a  semicircle  intersecting  CD  at 
E  and  F  respectively.  DE A  =90°  and  DFA  =  90°  (angles 
inscribed  in  a  semicircle). 

On  DC  determine  a  unit's  distance  S  =  DG;  draw  DJ 
±DC;  then  GH  and  GJ  are  the  two  values  of  x  which 
satisfy  the  equation,  and  in  this  case  they  are  both  negative 
answers.     The  proof  follows : 

GH  = 
1 


(12) 


-  tan  (b  =  -  tan  EDC 


(13)     tan#DC  = 


CE 
DC 


CE 


x. 


Cons.  GH  and   def. 
tangent. 


(14)  CE=-ax.  Mult,  in  (13) 

(15)  BE=BC-CE  =  b-(-ax)=b+ax  Sum  of  parts  Ax 

(16)  tan  A#£  =  tan  EDC=+x 

=  ~BE=~ll~E'  Def-tangent 


THE  QUADRATIC  EQUATION  253 

(17)  x(BE)  m  -c  Mult,  in  (16) 

(18)  x(ax+b)  =  -c  Sub.in(17)from(15) 

(19)  ax2+bx=-c  (18) 

(20)  ax2+bx+c  =  0  Trans.  (19) 

Therefore  x  satisfies  the  general  quadratic  equation  (3). 
Apply  this  method  to  examples  (1)  .  .  .   (5) 

Ex.7.    E2=E1z+E2*. 

Solve  for  E,  Ex,  and  E2. 

Ex2        E 


Ex.  8. 


Ex.9.     tt 


r22+z22     Xi 
Solve  for  r2  and  x2. 
T 


Solve  for  L  and  C. 


Ex.  10.    E=lVR2  +  <»2L2. 

Solve  for  72,  o>,  and  L. 

W 
Ex.11.    t  = —  sin  (td  —  6). 

Solve  for  #  and  L. 

7.  In  removing  a  radical  from  an  equation  the  latter 
should  be  transformed  so  as  to  place  the  radical  alone  in 
one  member.  Then  the  equation  should  be  raised  to  a 
power  corresponding  to  the  index  of  the  root. 

Ex.12.    /32=722+^+/l2. 

Solve  for  I3,  I2)  Iu 
E 


Ex.  13.    I  = 


>M^-£) 


Solve  for  R,  o>,  L,  C. 


254  PRACTICAL  MATHEMATICS 

In  order  to  solve  for  g>  proceed  as  follows:  Square  both 
members  of  the  equation  and  get  the  ( )  in  the  left  member 
alone.  Take  the  root  of  both  members  and  clear  of  frac- 
tions.    Solve  by  five-step  method. 

Observation.  When  an  unknown  quantity  is  involved 
under  several  indicated  signs  of  operations  it  can  be  freed 
therefrom  only  by  performing  the  indicated  operation  on  the 
quantity  or  by  performing  the  inverse  operation  upon  the 
entire  equation. 

Ex.  14.    E  =nLi  +jtoLJi  + 


Ex.  15.    E, 


r2+ju>L2' 
Solve  for  M  and  o>,  substitute  j2  =  —  1. 

E(rl2+x12) 


(ri+r2)2+(a:i+x2)2 
Solve  for  rh  r2,  xi,  x2. 


Ex.  16.    0i  = 


n2+zi2 
Solve  for  n  and  Xu 

Ex.17,     fr-     ^    .. 

ri2+X!2 

Solve  for  n  and  xi. 
Ex.18,    /,-r^--/^ 


Ex.  19. 


r22+x22    "r22+x22 
Solve  for  r2  and  x2. 

I2R_I2(Rr-Xx) 

r  r2-\-x2 

Solve  for  r  and  x. 


El.20.  P=«jf±p,w. 

r(Rr  —  Xx) 
Solve  for  r  and  x. 


THE  QUADRATIC  EQUATION  255 


re  for . 


Ex.  21. 

Solve  for  Nx  and  N. 


)X. 


Solve  for  Nl  and  iV. 

Ex.23.    W%-aVf&*. 

Solve  for  5. 

Ex.24.     JFC  =#'(/')  2+^2- 

Ex  r2w2M2        .  f    T        »*i*M» 

Ex.  25.    ^=n  +    ',     or,+J    coLx 


/i  r22  +  w2L22       I  r22  +  o>2L22 

Solve  for  L2,  M2,  r2. 


1  /2V'\ 2 
Ex.26.    R'-t{W)R. 

Solve  for  JV7  and  N. 

Lm  ->-*P(f44 

Solve  for  HP. 
28.    r--Bfafi+^J. 


Ex 


Ex.  29. 


Solve  for  a. 

E  —a-ar2Y 


Ex.  30.    s  - 


W    a*+a*ZYx+ZY 
Solve  for  a. 
agu—abv 


Ex.31.     f- 


tt2+#2 
Solve  for  u  and  v. 
abu+agv 


u2+v2 
Solve  for  u  and  v. 


256  PRACTICAL  MATHEMATICS 

A>+B>+2ABcos* 

Solve  for  A,  £,  #,  w,  and  L. 

8.  An  equation  is  homogeneous  for  two  or  more  of  its 
elements  if  they  are  interchangeable  without  altering  the 
value  of  the  equation.  If  the  value  of  one  element  is 
known  its  homogeneous  element  may  be  obtained  by  an 
interchange  of  letters. 

*  B*R+ABR  cos  b+ABaL  sin  <j> 

Solve  for  B,  R,  to,  L. 

Ex.34.     ,.J=gffi±sS3. 

pABL 

Solve  for  t,  R,  w,  L. 

Ex.35.   A2  =  (E+#/)2  +  g>2L2/2. 
Solve  for  A,  w,  L,  L 

P      AB  cos  (<]>-- 6) +#2  cose 
X*      *     P'  ~A£  cos  (<J>+8)  +A*  cos  6' 
Solve  for  A  and  5. 


Ex.  37.    B  =  VA 2  +#2  -  2A#  cos  6. 
Solve  for  A  and  #. 


„      no      „„  P2cos8-AP 

EX.38.      P^max  — 


VjB2  +  to2L2 

Solve  for  #,  w,  L,  B. 

9.  In  the  solution  of  any  given  equation  involving  long 
or  cumbersome  phrases  or  expressions,  substitute  a  single 
abbreviation  for  the  latter.  In  the  final  form  of  trans- 
formation reinstate  the  substituted  value. 


Ex.  39. 


THE  QUADRATIC  EQUATION  257 

AB  B2  cos  6 


Solve  for  B. 


Ex.  40.  A 2  =  (rI  +  y  V + (X  -  (oL7) ». 

Solve  for  «,  P",  X,  <o,  L. 
Ex.41.     T        0(n-*° 


l+6(n-n')2 
Solve  for  n  and  n'. 

10.  Parenthetical  quantities  may  be  solved  as  a  single 
unknown  in  equations  like  (41)  and  the  values  of  the 
enclosed  letters  determined  by  subsequent  transposition. 


Ex.  42. 

\    u2+v2 

Solve  for  a,  s,  E',  u,  and  v. 

Ex.  43. 

a     Z      aY1Z' 

s    ar-z        s 

Solve  for  a. 

Ex.  44. 

u  =  a2r2  +sr2  +a2r2(g1n  +hx) 

Solve  for  a. 

Ex.  45. 

p    sO-s)c(E')2 
d+hs+ks2  ' 

Solve  for  s. 

Ex.  46. 

T         sc(E')2 
d+hs+KS2 

Solve  for  & 

Ex.  47. 

T               Qr*s(E')2 

r22+2nr2s  +  {n2+x2)s2 
Solve  for  r2)  s,  n,  E' . 


258  PKACTICAL  MATHEMATICS 

E,48.  r.-   ^y* 

Solve  for  n. 

2r2(#')2 

Ex  49      T  = 2V     y 

•      •       °    r22+2nr2+n*+x2' 

Solve  for  n,  r2. 

„     M           \/#o-(#sin0  +  :ri)2  -E  cos-  0 
Ex.  50.    r  = z . 


Solve  for  E,  x,  I. 


Ex.  51.  In  (50)  substitute  cos  0  =  Vl  -  sin2  0  and  solve  for 
sin  0. 

Ex.  52.  Make  a  complete  list  of  all  quadratic  equations 
occurring  in  the  preceding  chapters,  solve  and  interpret  them. 

Ex.  53.  Make  a  complete  list  of  all  quadratic  and  biquadratic 
equations  occurring  in  the  following  chapters;  solve  and  interpret 
them. 

Ex.  54.  In  Ex.  33  substitute  wL  =  0  and  solve  for  B  and  R. 

Ex.  55.  In  Ex.  49  substitute  n  =  r2  and  solve  for  r2. 

Ex.  56.  In  Ex.  50  substitute  0  =  0°  and  solve  for  E. 

Ex.  57.  In  Ex.  50  substitute  0=90°  and  solve  for  E. 


CHAPTER  XIII 
THE  ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS 

The  elements  of  the  strength  of  materials  comprehend 
the  working  formulas  and  their  interpretations  when  applied 
in  the  design  of  structural  material.  One  purpose  is  the 
determination  of  the  proper  sizes  and  forms  to  withstand 
given  loads  and,  secondly,  to  determine  the  loads  which 
can  be  applied  safely  to  constructive  material  of  known 
dimensions.  A  third  purpose  is  the  acquisition  of  sufficient 
knowledge  for  the  use  of  a  handbook  of  materials. 

1.  Stress  per  Unit  of  Area. — The  total  load  or  weight  (P) 
in  pounds  upon  a  body,  is  the  product  of  the  area  (A)  in 
square  inches  of  the  resisting  surface  times  the  intensity  of 
stress,  i.e.,  the  stress  S  per  unit  of  area,  as  expressed  in  (1) 
and  (2).  This  is  the  fundamental  formula  for  direct  stresses, 
i.e.,  for  tension  and  compression. 

(1)  Load  =  unit  stress  X  area. 

(2)  P  =  SA. 

Solve  (2)  for  S  and  A,  and  interpret  the  resulting  equa- 
tions. 

Ex.  .1.     Determine  the  value   of  S,  when  P=  20000  lbs.  and 

A  =  2  sq.in. 

Ex.  2.  Determine  the  allowable  load  for  a  cross-section  of 
4.5  sq.in.,  when  the  unit  stress  is  2500  lbs.  per  square  inch. 

Ex.  3.  What  cross-section  is  required  to  sustain  a  load  of 
34000  lbs.,  when  the  allowable  unit  stress  =8000  lbs.  per  square 
inch? 

259 


260  PEACTICAL  MATHEMATICS 

Ex.  4.  What  is  the  dimension  of  the  side  of  a  square  oak 
timber  which  is  to  carry  a  compression  load  of  75000  lbs.,  allowing 
a  unit  stress  of  1500  lbs.  per  square  inch.  What  will  be  the  corre- 
sponding cross-sections  for  a  unit  stress  of  one-fifth  of  the  ultimate 
strength,  when  square  bars  of  iron,  wrought  iron,  and  steel  are 
substituted. 

2.  The  strength  of  any  material  is  determined  by  placing 
a  sample  of  the  material  in  a  testing  machine  in  which  it 
may  be  subjected  to  a  direct  stress  of  tension  or  com- 
pression. Under  these  respective  stresses  the  sample  is 
elongated  or  shortened,  and  the  deformation  is  propor- 
tional to  the  stress  until  the  elastic  limit  is  reached.  The 
elastic  limit  is  a  unit  stress  beyond  which  the  material  shows 
a  marked  permanent  set,  i.e.,  deformation.  The  ultimate 
strength  of  the  sample  is  the  highest  value  of  the  unit 
stress  just  before  the  sample  ruptures. 

The  modulus  or  coefficient  of  elasticity  (E)  is  the  ratio 
of  the  unit  stress  to  the  unit  deformation  (s)  as  expressed 
in  (3). 

(3)  E  =  -. 

The  unit  deformation  (s)  is  the  ratio  of  the  total  deforma- 
tion (e)  of  a  bar  to  its  length  (I)  as  expressed  in  (4). 

(4)  .4 

Substitute  in  (3)  the  value  of  S  from  (2)  and  the  value 
of  s  from  (4),  and  simplify.  The  values  of  the  coefficients 
of  elasticity  are  practically  equal  for  both  tension  and 
compression. 

The  materials  of  construction  may  be  ruptured  not  only 
by  tension  and  compression  but  also  by  shearing,  i.e.,  by 
transverse  breaking  or  cutting  along  ( a  section.  A  shear 
results  when  the  difference  in  the  forces  on  the  two  sides 
of  a  section  becomes  excessive. 


ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS    261 

The  materials  which  enter  into  the  construction  of 
buildings,  bridges,  and  machines  are  subject  to  various  com- 
binations of  the  elementary  stresses  of  tension,  compression, 
and  shear.  These  stresses  may  be  studied  by  their  effects, 
in  direct  deformation  by  compression  and  elongation,  and 
in  indirect  deformation  by  bending  and  twisting. 

The  values  of  the  strength  of  materials  vary  through  a 
wide  range,  depending  upon  the  quality  of  the  manufac- 
tured product.  For  any  specific  product  the  student 
should  consult  a  manufacturer's  handbook.  Table  X 
is  provided  as  a  guide  and  represents  a  compilation  from 
various  sources.  The  average  values  are  those  which  have 
been  accepted  in  practice. 

3.  A  beam  is  a  bar  of  rigid  material  which  is  supported 
in  some  manner,  at  one  or  more  points.  The  sum  of  the 
reactions,  i.e.,  the  forces  acting  upward  at  the  supports 
equals  the  sum  of  the  forces  acting  downward  upon  the 
beam. 

Ex.  5.  Make  a  drawing  of  a  beam  supported  at  two  points  Rt 
and  R2  and  separated  by  a  distance  d  feet.  At  a  point  P  on  the 
beam  distant  x  feet  from  Rlf  represent  a  concentrated  load,  i.e.,  a 
force  acting  downward.  The  beam  may  be  regarded  as  a  balanced 
lever  in  which  either  of  the  supports  acts  as  a  fulcrum.  Under 
this  consideration  the  product  of  the  load  times  its  lever  arm 
equals  the  reaction  times  its  lever  arm.  If  Ri  and  R2  also  designate 
the  respective  reactions  and  P  the  load,  then  (5)  expresses  the 
equilibrium  about  Ri  as  a  fulcrum,  and  (6)  expresses  the  equilib- 
rium about  R2  as  a  fulcrum. 

(5)  Bx=R2d. 

(6)  P(d-x)=Rld. 

(7)  P=R,+R2. 

(7)  results  by  adding  (3)  and  (4)  and  dividing  by  d.     Interpret  (7). 

4.  The  product  of  a  force  or  load  times  its  lever  arm  is 
called  a  moment.  The  weight  of  a  beam  or  a  uniformly 
distributed  load  is  equivalent  to  a  concentrated  load  at  the 


262 


PRACTICAL  MATHEMATICS 


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ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS    263 

center  of  weight.  The  sum  of  the  loads  equals  the  sum  of 
the  reactions.  If  loads  are  distributed  along  a  beam,  then 
the  sum  of  the  moments  about  one  support  equals  the  moment 
of  the  reaction  from  the  other  support. 

Ex.  6.  A  beam  whose  weight  is  35  lbs.  per  linear  foot  rests 
upon  supports  18  ft.  apart.  A  weight  of  400  lbs.  is  placed  at  5  ft. 
from  the  left  end,  and  a  second  weight  of  600  lbs.  at  7  ft.  from 
the  right  end.  Compute  the  reactions  due  to  the  separate  loads 
and  also  due  to  the  total  load. 

Ex.  7.  Compute  the  bending  moments  of  a  10-ft.  girder 
weighing  100  lbs.  per  linear  yard,  with  a  load  of  150  lbs.  at  the 
center.  Draw  the  diagram  of  bending  moments  and  repeat  after 
shifting  the  load  2  and  4  ft.  from  the  center.  The  bending  moments 
at  the  different  sections  may  be  represented  by  a  continuous 
series  of  vertical  lines  drawn  to  scale. 

5.  Any  point  on  a  beam  may  be  considered  as  a  fulcrum. 
The  algebraic  sum  of  the  load  moments  and  the  reaction 
moments  on  either  side  of  a  section,  is  called  the  bending 
moment  (M)  of  the  section  at  the  point  considered.  The 
forces  tending  to  cause  rotation  in  clockwise  direction  are 
positive,  while  those  tending  to  cause  counter-clockwise 
rotation  are  negative. 

Ex.  8.    Show  that   (8)   expresses  the  bending  moment  of  a 
uniformly  loaded  beam  of  length  /,  where  x  is  the  distance  of  the 
section  from  one  of  the  supports,  and  w  is  the  load  per  foot  of 
length. 
(8)  M  =  ±wx(l-x). 

What  is  the  value  of  the  bending  moment  for  a  mid  section, 
also  for  a  section  over  either  support.  Does  this  tendency  to 
rotation  indicate  the  bending  of  the  beam  downward? 
M*  A  parabola  whose  chord  is  I  in  length  and  whose  height  is 
\wl2  serves  as  a  diagram  for  the  graphic  representation  of  the 
varying  bending  moment.  The  bending  moment  at  any  section 
is  the  ordinate  of  the  parabola. 

6.  A  cantilever  is  a  beam  which  is  free  at  one  end.  In 
Fig.  102  the  free  end  is  at  the  left  whereas  the  right  end  is 


264 


PEACTICAL  MATHEMATICS 


supported  by  a  brick  wall.  There  is  no  bending  at  the 
left  end  and  since  the  moments  are  always  negative  the 
tendency  to  rotation  causes  the  bending  of  the  beam  at 
the  end.  (7)  is  the  equation  of  the  bending  moment  for  a 
cantilever  of  length  x.    P  is  the  concentrated  load  at  the 


Fig.  102.— A  Cantilever. 


left  end,  and  w  the  uniform  load  per  foot  of  length,  x\  is 
the  distance  of  the  cross-section  from  the  left  end. 


(9) 


M 


(9)  reduces  to  (10)  which  expresses  the  bending  moment 
at  the  support  for  a  cantilever  with  a  uniform  load  only. 


(10) 


M=- 


WX\' 


A  parabola,  shown  in  Fig.  20,  whose  half  chord  is  x  in 

wx^ 
length,  and  whose  height  is  — -,  is  a  diagram  which  repre- 

z 

sents  the  varying  bending  moment.  The  bending  moment 
is  greatest  at  the  wall  and  is  represented  as  the  largest 
ordinate  at  the  left  end  of  the  moment  diagram. 

7.  Center  of  Gravity.  The  centroid,  i.e.,  the  center  of 
gravity  of  a  mass  or  center  of  area  of  a  plane  figure,  may  be 
determined  experimentally,  graphically  or  by  computation. 
In  all  cases  it  is  the  point  about  which  the  material  of  the 
figure  is  equably  distributed.     The  sum  of  the  moments  of 


ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS    265 

the  gravitational  forces  acting  on  the  elements  of  the  figure 
on  one  side  of  the  center  of  gravity,  equals  the  sum  of  the 
moments  of  the  gravitational  forces  acting  on  the  other 
side. 

Ex.  9.  Make  a  duplicate  of  Fig.  103  on  a  heavy  piece  of  card- 
board. Place  the  figure  on  a  knife  edge  or  table  edge,  and  when 
balanced  draw  a  line  on  the  cardboard  above  the  balancing  edge. 
Balance  the  cardboard  in  a  new  position,  and  draw  a  line  over 
the  balancing  edge.  The  intersection  of  these  two  lines  is  the 
centroid  of  the  figure.     Repeat  for  a  new  position  as  a  check. 

For  figures  of  symmetry  the  centroid  lies  at  the  intersection 
of  the  axes  of  symmetry. 

Ex.  10.  Make  a  tracing  of  Fig.  103  on  cross-section  paper  so 
that  the  line  XX',  i.e.,  the  axis  of  reference,  is  horizontal.  The 


Fig.  103. — The  Centroid  of  a  Figure. 

horizontal  ruling  of  the  paper  divides  the  figure  into  thin  elements, 
i.e.,.  slices  of  approximately  rectangular  areas.  The  moment  of 
any  one  of  these  rectangles  about  the  axis' XX',  is  its  area  times  its 
mean  distance.  The  sum  of  the  moments  of  the  rectangular 
areas  equals  the  moment  of  the  whole  area.  Designating  the 
elementary  areas  by  bx,  b2,  b3,  .  .  .  ,  and  their  moment  arms  by 
Vh  2/2,  2/3,  ...  ,  respectively,  and  the  whole  area  by  A  and  its 
moment  arm  by  z,  (11)  states  that  the  sum  of  the  moments  of 
the  elementary  areas  equals  the  moment  of  the  whole  area. 


(11) 


biyi  +622/2  -f&3?/3+.  .  .=Az. 


266  PRACTICAL  MATHEMATICS 

Abbreviate  biyi+b2y2+b3yz+.  .  .  by  iby,  in  which  s  is  a 
symbol  of  summation  of  all  products  of  a  b  term  times  a  y  term. 
Therefore  (11)  may  be  written  (12). 

(12)  xby  =Az. 

8.  The  interpretation  of  (13)  states  that  the  moment  arm 
(z),  of  the  whole  figure  (A)  about  the  axis  XX',[is  the  ratio 
of  the  sum  of  the  moments  of  the  elementary  areas  to  the 
whole  area.  If  the  entire  area  were  concentrated  at  the 
distance  z  from  XX'  the  moment  of  the  figure  would 
remain  unchanged  in  value.  The  centroid  of  Fig.  103 
is  located  at  a  distance  z  from  the  XX'  axis.  The  distance 
of  the  centroid  from  a  vertical  axis  of  reference  may  be 
obtained  in  a  similar  manner  by  dividing  the  figure  into 
vertical  strips.  The  ratio  of  the  sum  of  the  moments  of 
the  new  areas  about  the  vertical  axis  of  reference,  to  the 
whole  area  is  the  corresponding  distance  of  the  centroid 
from  the  vertical  axis.  Draw  two  lines  parallel  to  the  two 
axes  at  distances  to  correspond  to  the  z  values.  Their  inter- 
section locates  the  centroid. 

Make  the  necessary  measurements  for  the  area  and 
moment  arm  of  each  strip  both  for  the  vertical  and  hori- 
zontal axes  of  reference,  and  compute  the  position  of  the 
centroid  of  the  figure.  The  accuracy  of  the  result  in  Ex.  10, 
will  increase  as  the  widths  of  the  areas  decrease.  A  more 
accurate  value  would  be  obtained  if  the  figure  were  divided 
into  elementary  squares  a\,  «2,  0,3,  •  •  •  ,  and  then  the  ratio 
of  the  sum  of  the  moments  of  the  elementary  figure,  to 
the  whole  area  will  give  the  axal  distance  of  the  centroid. 
Suppose  Fig.  103  is  rotated  about  the  axis  XX'  generating 
a  ring  of  volume  (V),  then  by  the  Theorem  of  Guldinus  we 
obtain  (14). 

(14)  V=2kzA; 


ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS    267 

z  is  the  axal   distance  of  the  centroid  of  the  cross-sectional 
area  A . 

Each  unit  of  area  a\,  a%,  as,  etc.,  in  its  rotation  generates 
a  part  of  the  ring  at  the  respective  axal  distance  of  2/1, 
2/2,  2/3,  etc.  Therefore  the  volume  (V)  equals  the  sum  of 
the  volumes  of  the  elementary  rings.  In  (15),  2%  may  be 
written  outside  the  summation  symbol  because  it  is  a 
common  factor  in  every  term. 

(15)  V  =  ^2%ay  =  2^ay, 
and 

(16)  2^ay  =  2%zA. 

(17)  .'.     z^-f. 

The  interpretation  of  (17)  is  identical  with  the  inter- 
pretation of  (13). 

Ex.  11.  Construct  the  following  figures  on  cross-section  card- 
board, and  determine  center  of  gravity  of  each:  (a)  a  semicircle; 
(b)  a  T  section;  (c)  an  I  section;  (d)  a  channel;  (e)  a  rail  section. 

9.  The  second  moment  of  an  elementary  mass  or  an 
elementary  area,  is  the  respective  mass  or  area  multiplied 
by  the  square  of  its  distance  from  an  axis  of  reference. 
The  sum  of  the  second  moments  of  the  elementary  masses 
or  areas  of  a  figure,  is  called  the  moment  of  inertia  (/) 
of  the  mass  or  area.  In  Fig.  103  the  elementary  areas  are 
squares.  If  a\,  a2,  as,  ...  ,  are  the  elementary  areas  and 
2/i,  2/2,  2/3?  •  •  •  >  their  respective  axal  distances  then  the 
moment  of  inertia  (J)  is  expressed  in  (18). 

(18)  I  =  2ay2. 

Ex.  12.  Determine  the  moments  of  inertia  for  the  figures 
used  in  Ex.  10. 


268  PEACTICAL  MATHEMATICS 

10.  The  ratio  of  the  moment  of  inertia  of  a  mass  or  an 
area,  to  its  respective  mass  or  area,  expresses  the  square 
of  the  axal  distance  (K)  at  which  the  material  could  be 
concentrated  in  order  to  retain  the  same  value  for  its 
moment  of  inertia.  K  is  also  called  the  radius  of  gyration 
as  expressed  in  (17)  when  J  is  the  least  moment  of  inertia 
of  the  figure. 

(19)  K=^A 

Solve  (19)  for  I  and  interpret. 

11.  The  elements  entering  into  a  moment  are  the  area 
and  the  arm,  and  since  an  area  varies  as  the  square  of  a 
linear  unit,  and  the  arm  varies  as  the  first  power  of  a 
linear  unit,  then  the  product  varies  as  the  third  power  of  a 
linear  unit.  If  the  linear  unit  is  one  inch  then  the  unit  of 
a  moment  is  a  cubic  inch.  The  moment  of  inertia  is  the 
product  of  an  area  times  the  square  of  the  arm  and  there- 
fore a  moment  of  inertia  is  measured  in  biquadratic  inches, 
i.e.,  the  fourth  power  of  a  linear  unit. 

12.  The  fibers  of  a  loaded  beam  are  subjected  to  tension 
on  one  side  and  compression  on  the  other  side.  There  is  a 
neutral  axis,  i.e.,  a  line  in  the  section  of  the  beam,  at  which 
the  stress  is  zero.  The  neutral  axis  passes  through  the 
center  of  gravity  of  the  section  and  is  the  axis  about  which 
the  sum  of  the  moments  of  the  internal  forces  equals  zero. 
The  stress  (S)  in  any  fiber  varies  as  its  distance  (c)  from 
the  neutral  axis  *Socc. 

«  •••  I,-;- 

In  (20)  S  is  the  stress  in  the  most  remote  fiber,  and  c 
is  its  corresponding  distance  from  the  neutral  axis.  Si  is 
the  stress  in  a  fiber  at  a  distance  y  from  the  neutral  axis. 

The  algebraic  sum  of  the  moments  of  the  internal 
horizontal  stresses  about  any  point  in  the  section  is  called 


ELEMENTS  OF  THE  STEENGTH  OF  MATERIALS    269 

the  resisting  moment  (ft).  If  the  bending  moment  be  com- 
puted, about  the  same  point  in  the  section,  then  the  bending 
moment  equals  the  resisting  moment  (21). 

(21)  R=M. 

The  unit  stress  may  be  calculated  by  dividing  the  stress 
($)  in  the  outside  fiber  by  its  distance  c  from  the  neutral 
axis.  Solving  (20)  we  obtain  (22)  the  expression  for  the 
stress  S\  of  a  unit  area  at  the  distance  y\. 

(22)  &-!*• 

Interpret  (22)  in  terms  of  unit  stress. 

If  the  fiber  has  an  area  a  then  (22)  becomes  (23). 

(23)  aSi=-ayi. 

c 

Interpret  (23). 

(24)  /.     R=2—y-y=-2ay2.  Def .  of  R 

c  o 

S 

—  is  a  constant  factor  in  all  the  terms  of  the  summation 

c 

and  therefore  may  be  written  preceding  the  summation 
symbol. 


Def.  of/ 

Subs,  in  (24) 

Subs,  in  (26)  from  (21) 


Interpret  (24)  and  (27) ;   solve  (27)  for  S,  c,  and  /  and 
interpret  the  resulting  equations. 


But 

(25) 

Xay2=I. 

(26) 

c 

(27) 

.-.    M--7. 

c 

270 


PRACTICAL  MATHEMATICS 


The  moments  of  inertia  for  standard  sections  are  given 
in  manufacturer's  tables. 

The  moment  of  inertia  (7)  of  a  rectangular  cross-section 
of  breadth  (6)  and  depth  (d)  is  expressed  in  (28)  when 
the  neutral  axis  passes  through  the  center  of  gravity  and 
in  (29)  when  the  axis  is  the  lower  edge  of  the  rectangle. 


(28)     t-%. 


(29)     1  = 


bd3 


Interpret  (28)  and  (29). 

The  moment  of  inertia  of  a  hollow  rectangle  (I*),  may 
be  obtained  by  subtracting  the  moment  of  inertia  (7i)  of 
the  interior  rectangle,  from  the  moment  of  inertia  (I)  of 
the  outside  rectangle. 


(30) 


h  =  I 


h^(bd» 


6idi3). 


Ex.  13.  Fig.  104  illustrates  a  hollow  rectangular  section 
turned  over  on  its  side.  Compare  the  moment  of  inertia  of  the 
section  in  this  position  with  the  moment  of  inertia  of  the  section 
in  its  normal  position,     b  =4  ins.,  6i  =3  ins.,  d  =8  ins.,  di  =7  ins. 


it 

.    6  it    . 

Ni 

., — b — > 

*iF 

1 

t 

- 

&i 

L 

Fig.  104. 


Fig.  105.— I-Beam. 


13.  The  moment  of  inertia  of  an  I-beam  section,  Fig.  105, 
may  be  computed  from  (28)  by  subtracting  the  moment 
of  inertia  of  the  two  vacant  rectangular  spaces  from  the 
moment  of  inertia  of  the  circumscribing  rectangle  as  ex- 
pressed in  (31). 


(31) 


7=^S6d3-(6-i)(d-2<)3}. 


ELEMENTS  OF  THE  STEENGTH  OF  MATERIALS    271 

Ex.  14.  Fig.  106  illustrates  a  T-beam  section.  By  the  use 
of  (29)  and  the  principle  of  Ex.  13,  we  obtain  (32),  where  Ci  and  c 
are  the  respective  distances  of  the  outside  fibers  from  the  neutral 
axis. 

(32)  /     l{fc3+6ci3_(6_£)(ci_£i)3). 


« 


-4 


t 

"     2      ~* 

'.1 

I 

t> 

t. 

\   , 

Fig.  106.— T  Beam. 


Fig.  107.— Channel. 


Ex.  15.  Fig.  107  illustrates  a  channel  section.  Show  that  (32) 
is  also  the  formula  for  the  moment  of  inertia  of  a  channel. 

Ex.  16.  Determine  the  area  and  the  moment  of  inertia  of  a 
standard  I-beam  with  the  following  dimensions  d  =  18  ins.,  6=6  ins., 
t  =  A6  in.  The  dimensions  of  the  flange  are  taken  from  a  manu- 
facturer's table. 

Ex.  17.  Determine  the  area  and  the  moment  of  inertia  of  a 
standard  T-bar  with  the  following  dimensions:  6=4  ins.,  thickness 
of  flange  U  =  |  to  ts  ins.,  thickness  of  stem  t  =  §  to  A  ins.,  depth 
c  =2.82  ins.,  and  ci  =  1.18  ins.     Check  with  a  manufacturer's  table. 

Ex.  18.  Determine  the  area  and  the  moment  of  inertia  of  a 
standard  channel  with  the  following  dimensions:  6  =  15  ins.,  Ci  =  .79 
in.,  c  =2.61  ins.,  t  =  .4  in.,  and  the  average  thickness  of  flange  =  .4  in. 
Check  with  a  manufacturer's  table. 

Ex.  19.  Make  a  drawing  from  the  detail  measurement  of  the 
section  of  a  standard  100-lb.  T-rail.  Determine  its  area  and 
moment  of  inertia  and  check  with  a  manufacturer's  table. 

14.  Comparative  Strengths  of  Beams  of  Equal  Length 
and  of  Equal  Cross-section.    From  (27)  we  obtain  (33). 


(33) 


S='-jM. 


SozM. 


The  ratio  -  is  called  the  section  modulus,  and  contains 
c 

all  the  dimensions  of  the  cross-section.     Therefore  for  a 


272  PRACTICAL  MATHEMATICS 

constant  section  modulus  the  stress  in  the  outside  fiber 
will  vary  directly  as  the  bending  moment.  If  a  number 
of  beams  are  compared  for  strength,  we  shall  assume  that 
they  have  the  same  section  modulus  and  the  same  unit 
stress,  and  therefore  the  bending  moments  will  be  equal 
by  (27).  Consider  four  beams  of  equal  cross-section: 
(1)  A  cantilever  loaded  at  the  end  with  Wi;  (2)  a  cantilever 
loaded  uniformly  with  W2;  (3)  a  simple  beam  loaded  at 
the  middle  with  W3;  and  (4)  a  simple  beam  loaded  uni- 
formly with  W±.  The  respective  bending  moments  are 
Mh  M2,  M3,  M4.    We  then  have  (34). 

(34)  M1  =  M2  =  Mb  =  Ma. 
For  beams  of  equal  length  I,  we  obtain 

(35)  iti-Wtl,    M2  =  Wf,    M*=?g,    M^f 

(36)  ,     Wll=Y  =  Y=lT- 

(37)  ,   ri-^J^ 

The  interpretation  of  (37)  states  that  the  uniformly  loaded 
simple  beam  mil  sustain  twice  the  load  of  a  simple  beam 
loaded  at  the  middle,  and  the  latter  will  sustain  twice  the  load 
of  a  cantilever  loaded  uniformly  and  again  the  last  will  sustain 
twice  the  load  of  a  cantilever  loaded  at  the  end.  Therefore, 
the  relative  strength  of  the  four  beams  in  the  order  mentioned 
isS  :4  :2  : 1. 

Ex.  20.  A  uniformly  loaded  standard  I-beam  has  its  supports 
20  ft.  distant.  The  beam  is  a  6-in.  B  17  Cambria  beam.  Deter- 
mine the  safe  load  allowing  a  fiber  stress  of  12,500  lbs.  Determine 
the  safe  load  if  the  beam  is  used  as  a  cantilever.  Determine  the 
equivalent  Carnegie  beam. 


ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS     273 

15.  The  stiffness  of  a  beam  is  the  ratio  of  its  deflection 
to  its  span.  Its  deflection  is  the  lateral  displacement  from 
its  normal  position. 

16.  Columns  and  struts  are  subjected  to  a  direct  com- 
pression stress  due  to  a  load  P,  which  causes  unit  stress  Sd, 
as  expressed  in  (1)  and  (2),  and  also  to  a  stress  Si  due  to 
bending.  The  total  stress  S  is  the  sum  of  the  contributing 
stresses  as  stated  in  (38). 

(38)  S  =  Sa+Si. 

(39)  S=-+C-^  Sub.  in  (38)  from  (2)  and  (27). 

The  only  force  causing  the  bending  moment  M  is  the 
load  P,  and  its  lever  arm  (/)  is  the  lateral  deflection  of  the 
center  line  of  the  column  from  its  normal  position. 

(40)  .\     M  =  Pf. 

(41)  I=Ar2. 

(41)  expresses  the  moment  of  inertia  of  the  cross- 
section  whose  area  is  A  and  whose  radius  of  gyration  is  r. 
Substituting  (40)  and  (41)  in  (39)  we  obtain  (42). 

PL  .  cf 


(42)  S  =  j{1+? 

Interpret  (42). 

The  deflection  of  a  beam  varies  directly  as  the  square 
of  its  length,  and  therefore  if  the  proportionality  factor  be 

taken  as  —  we  obtain  (43). 
c 

fozP;        /.     (43)    f=^P        and         /.     (44)     cf=ql2. 
c 

P        S 


(45) 


A     ^P- 


274  PRACTICAL  MATHEMATICS 


(45)  is  the  result  of  substituting  (44)  in  (42)  and  solving 
p 
for  — .     The  factor  q  is  a  numeric  constant,  which  depends 

upon  the  kind  of  material,  and  the  arrangement  of  the 
ends  of  the  columns.  Interpret  (45)  and  solve  for  P,  A, 
q,  I,  and  r,  and  interpret  each  of  the  resulting  equations. 

Ex.  21.     Determine   the   safe   load    on   a   column   for   which 

I2                              1 
S  =8000  lbs.  per  square  inch,  A  =3x4, —  =4800,  and  q= . 

T  oOOO 

17.  The  moment  of  inertia  (M)  of  a  circle  about  its 
diameter  (d)  is  expressed  in  (46). 

(«)  M=^. 

Ex.  22.  Divide  (46)  by  the  area  of  the  circle  and  determine 
the  radius  of  gyration  of  the  circle  referred  to  the  diameter. 

Ex.  23.  Construct  a  circle  with  two  perpendicular  diameters. 
The  radius  of  gyration  is  n.  Lay  off  the  two  lines  parallel  respect- 
ively to  the  perpendicular  diameter  and  at  distances  therefrom 
equal  to  the  radius  of  gyration.  The  intersection  of  these  lines 
determines  a  point,  whose  distance  from  the  center  is  r.  A  radius 
equal  to  r  is  called  the  polar  radius  of  gyration.  Show  that 
r2=2n2. 

18.  A  polar  moment  of  inertia  (J)  is  a  second  moment 
of  an  area  about  a  point  called  a  pole.  In  the  case  of  the 
circle  the  pole  is  the  center  of  the  circle.  The  polar  moment 
is  the  product  of  the  area  of  the  figure  times  the  square  of 
its  polar  radius  of  gyration  as  expressed  in  (47). 

(47)  J=Ar2. 

But 

A=-r-        and      ri=T       and      r2  =  2n2  =  ^-. 


ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS    275 

Therefore  the  polar  moment  of  inertia  is  twice  its  axal 
moment  of  inertia. 

Ex.  24.  The  polar  moment  Ja  of  an  annulus  whose  external 
and  internal  diameters  are  di  and  d2  respectively  is  given  in  (49). 

(49)  /a=^i4-44). 

Ex.  25.  Show  that  (49)  reduces  to  (50)  where  A  expresses 
the  area  of  the  annulus. 

(50)  Ja  = g • 

19.  The  polar  moment  of  inertia  of  any  section  is  the 
sum  of  the  moments  of  inertia  taken  with  respect  to  any 
two  perpendicular  axes. 

Ex.  26.  Round  Shaft  to  Transmit  Power. 

(51)  Ssd*  =321000-. 

n  -,.  _ 

Notation:  I 

H  =  horse-power  transmitted; 
n  =  number  revolutions  per  minute; 
d  =  diameter  in  inches; 
Ss  =unit  stress  for  shearing  per  square  inch. 

Solve  for  Ss,  d,  H,  and  n,  and  interpret  the  resulting  equation. 
Determine  the  unit  stress  in  a  shaft  from  the  following  data: 

#=25;      ft  =  100;      d  =2.5  ins. 
What  is  the  factor  of  safety  for  this  shaft? 

20.  Combined  Bending  and  Twisting.  When  a  shaft 
is  subject  to  both  bending  and  twisting  then  (52)  expresses 
the  unit  stress  Si  due  to  the  greatest  bending  stress  S 
and  the  tortional  shearing  unit  stress  Ss. 


(52)  •        Sl  =  S+Vf+S2 

API 

(53)  S=g. 


276  PRACTICAL  MATHEMATICS 

Substitute  in  (52)  the  value  of  S  from  (53),  and  the 
value  of  Ss  from  (51),  and  simplify.  P  represents  the  load 
on  the  shaft  and  I  the  length  of  the  shaft. 

The  last  equation  has  been  modified  by  A.  E.  Wiener, 
who  suggests  (54)  for  the  bearing  portion  of  a  shaft  of 
diameter  db,  and  (55)  for  the  core  portion  of  a  shaft  of 
diameter  dc,  in  which  W  is  the  capacity  of  the  generator 
in  watts,  fa  and  fa  are  constants,  fa  ranges  in  value  from 
.0025  for  a  high-speed  drum  armature  to  .005  for  a  low- 
speed  ring  armature,  fa  ranges  in  value  from  1  for  a  1-KW. 
machine  to  1.8  for  a  2000-KW.  machine. 

(54)  db  =  kypVn. 

(55)  dc  =  fayJ^ 

Ex.  27.  Determine  db  and  dc  for  a  10  KW.  machine  for  which 
?i  =  1700,  fo  =  .0025,  and  &c=1.2. 

21.  Pulley  Diameter.  The  diameter  dP  of  a  pulley  is 
expressed  in  (56)  in  which  v  is  the  velocity  of  the  belt  in 
feet  per  minute. 

(56)  dP=  — . 
Simplify  (56). 

22.  Armature  Bearings.  The  length  (I)  of  an  armature 
bearing  is  expressed  in  (57)  in  which  fa  is  a  constant  ranging 
from  .1  to  .225  for  high-speed  and  from  .15  to  .3  for  low- 
speed  armatures. 

(57)  l  =  fadbVn. 

Ex.  28.    Determine  I  for  Ex.  27  using  kt  =  .2. 

23.  Riveted  Joints.     Notation: 
p  =  pitch  of  rivets; 

d  =  diameter  of  rivets; 
t  =  thickness  of  plates; 


ELEMENTS  OF  THE  STRENGTH  OF  MATERIALS    277 

St= tensile  strength  or  resistance  of  plates  to  tension; 
Ss  =  shearing  strength  or  resistance  of  plates  to  shearing ; 
Sc  =  crushing  strength  or  resistance  of  plates  to  crush- 
ing. 

Considering  a  width  of  the  joint  equal  to  the  pitch  of 
the  rivets: 

(58)  The  resistance  of  this  portion  to  tearing    =  (p  —  d)tSt. 

(59)  The  resistance  of  this  portion  to  shearing  =  .7854<i2>Ss. 

(60)  The  resistance  of  this  portion  to  crushing  =  dtSc. 

(61)  The  resistance  of  the  solid  plate  to  tearing  =  ptSt. 

Ex.  29.  When  tearing  resistance  equals  shearing  resistance, 
then  (58)  =(59).  What  will  be  the  value  of  p?  What  will  be  the 
value  of  d? 

Ex.  30.  If  in  a  width  equal  to  p,  there  are  n  rivets  in  single 
shear  then 

(62)  (p-d)tSt=.785±d*nSs. 

Solve  for  p,  d,  t,  St,  Ss,  and  n.  Interpret  (62)  and  the  resulting 
equations. 

Ex.  31.    If  the  n  rivets  are  in  double  shear  then 

(63)  (p  -d)tSt  =  .7854c?2  2nSs. 

Solve  for  p,  d,  t,  n,  Sh  and  Ss.  Interpret  (63)  and  the  resulting 
equations. 

Ex.  32.  If  the  resistance  to  shearing  equals  the  crushing 
resistance,  then  (59)  =  (60).    What  will  be  the  value  of  d? 

Allow  Sc  =2SS  and  substitute  in  the  last  equation. 


24.  Efficiency  of  a  Joint.     The  efficiency  of  a  joint  is  the 

io  or  - — -  whichever  is  least.     Int< 

(61)        (61) 

and  express  the  ratio  as  efficiency  per  cent. 


ratio or  ; —  whichever  is  least.     Interpret  this  ratio 

(61)        (61) 


278  PRACTICAL  MATHEMATICS 


25.  Lap  Joints. 

(64)  For  single-riveted  lap  joints  pi  =  : 75 — -+d. 

(65)  For  double-riveted  lap  joints        p2  = 7^ r«. 

to* 

,*^     -n  vi     •     x  j  1      -'■."*  3X.7854d2&  .  . 

(66)  For  treble-riveted  lap  joints         ps  = -= \-d. 

(67)  For  quadruple-riveted  lap  joints  p±  = '—r^ -+d. 

Ex.  33.  If  rivets  of  like  quality  and  equal  dimensions  were 
used  for  these  four  cases  of  lap  joints,  what  would  be  the  relative 
pitches? 

26.  Butt  Joints. 

(68)  For  single-riveted  butt  joints       pi  = '—t~ — — -+d. 

(69)  For  double-riveted  butt  joints      p2  = L-r^ -+d. 

tbt 

(70)  For  treble-riveted  butt  joints        p3= '—r& s~\-d. 

tbt 

Ex.  34.  What  would  be  the  relative  pitches  for  the  three  cases 
of  butt  joints  if  rivets  of  like  quality  and  equal  dimensions  were 
used? 


CHAPTER  XIV 
THE  USE  OF  SQUARED   PAPER 

1.  The  employment  of  graphic  methods,  or  the  pictorial 
representation  of  the  magnitudes  of  related  facts  has  been 
adopted  in  every  sphere  of  human  activity,  both  for  expo- 
sition and  for  calculation. 

The  facts  of  life  are  recorded  in  many  ways.  When 
they  are  tabulated  they  are  called  statistics  or  data.  The 
results  of  experiments,  the  enumerations  of  newspapers, 
and  the  government  compilations  of  population,  production, 
wealth,  debt,  and  taxation  are  not  only  presented  in 
tabulated  form  but  are  expressed  graphically  in  the  form 
of  charts. 

Everything  in  the  universe  is  subject  to  change,  since 
both  animate  and  inanimate  objects  either  grow  or  modify, 
i.e.,  they  vary  with  a  change  in  time,  place,  and  stress. 

Our  position  in  space,  our  location  upon  the  earth  and 
the  situation  of  our  homes  are  expressions  of  relations, 
distances,  and  directions  between  objects. 

By  tabulating  or  grouping  facts  we  establish  an  intimate 
relation  between  at  least  two  variables,  i.e.,  two  kinds  of 
changing  quantities.  One  of  these  variables  is  often  the 
element  of  time. 

The  data  furnished  from  a  test  of  a  series  railway  motor 
are  recorded  in  Table  XI.  The  variables  are  speed  (R.P.M.) 
in  revolutions  per  minute  and  current  (I)  in  amperes.  In 
this  manner  we  place  together  a  collection  or  record  of 
measured  facts,  which  shows  the  change  of  current  in  the 
motor  with  a  corresponding  variation  in  its  speed. 

279 


280  PRACTICAL  MATHEMATICS 

TABLE  XL     TEST    UPON    A    SERIES    RAILWAY  MOTOR 


Point  in  Fig.  108.  . 

A 

B 

C 

D 

E 

F 

G 

H 

J 

K 

Speed  (R.P.M.)... 

2050 

1230 

950 

805 

705 

630 

575 

530 

495 

470 

Current  (amperes) 

15 

20 

25 

30 

35 

40 

45 

50 

55 

60 

By  noting  in  Table  XI  the  change  in  current  with  the 
corresponding  successive  values  in  speeds  we  can  form  an 
idea  of  the  influence  of  the  current  variations.  We  can 
appreciate  these  facts  if  the  eye  can  see  them  recorded 
with  smaller  intervals  of  change,  or  preferably  by  a  con- 
tinuous record.  This  may  be  accomplished  with  the  aid  of 
cross-section  paper,  as  shown  in  Fig.  108. 

2.  Cross-section  Paper.  Squared  cross-section  paper  is 
a  sheet  of  paper  which  has  been  ruled  carefully  with  hori- 
zontal and  vertical  lines.  These  are  spaced  equally,  and 
usually  there  are  10  or  20  divisions  to  the  inch.  Every 
fifth  or  tenth  line  is  accentuated  to  facilitate  the  counting 
of  spaces  and  the  avoidance  of  errors.  The  location  of  any 
point  on  the  paper  is  determined  by  its  proximity  to  the 
nearest  horizontal  and  the  nearest  vertical  line.  The 
bounding  lines  serve  as  beginning  lines  or  lines  of  reference 
from  which  all  other  lines  are  numbered  in  order.  The 
two  lines  of  reference  or  zero  lines  are  called  the  axes.  The 
position  of  a  point  is  determined  by  its  distance  from  the 
two  axes  of  reference,  and  accordingly  every  point  is 
designated  by  a  pair  of  values.  An  8.5"  X  5.5"  perforated 
paper  has  a  ruled  space  7.5"  X  5"  subdivided  into  twentieths 
of  an  inch.     Every  tenth  line  is  accentuated. 

3.  Plotting  Tabulated  Data.  Upon  the  cross-section 
paper  we  plot,  i.e.,  locate  a  series  of  points  corresponding  to 
each  pair  of  values  in  the  data  of  the  table.  A  smooth 
curve  is  drawn  through  the  plotted  points  with  the  aid  of 
a  flexible  rule  or  a  French  curve. 


THE  USE  OF  SQUARED  PAPER 


281 


The  resulting  curve  is  designated  in  terms  of  its  variables, 
and  is  described  as  a  locus  (position  of  points)  and  is  called 
a  graph  (a  writing  or  pictured  fact). 

A  pair  of  values  is  required  to  locate  each  point.  It  is 
necessary  to  choose  suitable  scales  to  be  applied  to  the 
axes  so  that  the  respective  values  of  the  variables  may  fall 
within  the  range  of  the  paper. 

Beginning  at  the  lower  left-hand  edge  of  the  cross- 
section  paper,  ink  over  the  accentuated  horizontal  and  ver- 
tical lines  bounding  the  paper.     (See  Fig.  108.) 

These  lines  are  called  the  axes  of  reference,  and  usually 
extend  beyond  their  intersection.  The  latter  is  called  the 
origin  and  is  the  zero  for  measurement  on  the  respective 
axes. 

On  the  two  axes  of  reference  lay  off  distances  to  suitable 
scales  so  as  to  comprehend  the  magnitudes  of  the  table. 


The    first    pair    of    values   to  be    plotted    is 


2050 


i:> 


and  the  last  pair  is 


K 


470 


60 


The  greatest  value  of  the  current  is  60.  There  are  100 
divisions  in  the  horizontal  direction  of  the  paper.  If  we 
lay  off  a  scale  for  amperes  on  the  horizontal  axis,  then  one 
division  on  the  paper  will  correspond  conveniently  to  one 
ampere.  It  will  require  60  divisions  for  60  amperes  and 
therefore  3  ins.  of  length  in  the  horizontal  direction  (assum- 
ing 20  divisions  per  inch) . 

Examining  the  table  again  we  notice  that  the  highest 
magnitude  for  speed  is  2050.  There  are  150  divisions  in  the 
vertical  direction   of  the  paper.     If  we   select  one  division 


282 


PRACTICAL  MATHEMATICS 


Fig.  108— The  Plotting  of  Tabulated  Data. 


THE  USE  OF  SQUARED  PAPER       283 

to  correspond  to  20  R.P.M.  the  paper  will  not  be  long 
enough.  Suppose  we  choose  one  division  to  correspond  to 
25  R.P.M.  Then  82  divisions  will  suffice  or  slightly  more 
than  4  ins.  of  length. 

Label  the  two  axes  Current  I  and  Speed  R.P.M.  respect- 
ively. At  convenient  intervals  on  the  axes  mark  and 
designate  units  of  division.  Every  5  amperes  and  every 
200  revolutions  should  suffice  to  aid  the  eye  in  reading  the 
scale. 

The  first  point  to  be  located  is  A,  which  corresponds  to 
the  first  pair  of  associated  values  in  Table  XI. 

On  the  horizontal  (I)  axis  locate  the  point  of  division 
corresponding  to  15  amperes.  Move  the  finger  upward 
over  the  vertical  line  passing  through  this  point.  Pause 
where  it  is  intersected  by  the  horizontal  line  passing  through 
2050  located  on  the  vertical  (R.P.M.)  axis.  Mark  this 
point  of  intersection  by  making  it  the  center  of  a  small 
circle  of  ^  in.  radius  and  label  it  A.  The  distance  measured 
from  the  vertical  axis  is  called  the  abscissa  of  the  point, 
whereas  the  distance  measured  from  the  horizontal  axis  is 
called  the  ordinate  of  the  point. 

The  second  point  to  be  plotted  is  B  which  corresponds 
to  the  second  pair  of  associated  values  in  the  table. 

Locate  20  on  the  /  axis  and  1230  on  the  R.P.M.  axis. 
The  intersection  of  the  vertical  and  horizontal  lines 
passing  through  these  respective  axal  divisions  is  the 
second  plot.  Indicate  it  with  a  circle  of  ^  in.  radius  and 
label  it  B. 

Proceed  in  like  manner  for  the  remaining  pairs  of  values. 
In  this  way  we  shall  have  plotted  ten  points:  A,  B,  C, 
D,  E,  F,  G,  H,  J,  K. 

A  smooth  curve  is  drawn  through  the  consecutive 
plots.  Every  point  on  the  curve  has  a  definite  pair  of 
values  for  its  /  and  R.P.M.  These  are  determined  by  the 
respective  horizontal  and  vertical  lines  passing  through  the 
specified  point  on  the  curve.     The  intersection  of  the  men- 


284 


PRACTICAL  MATHEMATICS 


tioned  lines  with  the  axes  gives  the  magnitude  in  amperes 
and  revolutions  per  minute  respectively. 

The  curve  is  a  succession  of  points.  It  is  for  this  reason 
that  the  curve  makes  a  continuous  record  and  therefore  the 
curve  will  supplement  the  values  given  in  Table  XI.  The 
amplification  of  the  table  by  means  of  the  curve  is  called 
interpolation. 

It  is  not  difficult,  although  not  always  advisable,  to  pro- 
ject the  curve  beyond  the  limiting  values  of  the  table.  Such 
an  extension  of  the  table  is  called  extrapolation.  This  is 
identical  with  the  process  by  which  the  human  mind  projects 
the  reason  and  predicts  with  prophetic  inspiration  future 
happenings  which  at  present  are  outside  of  the  realm  of 
recorded  knowledge. 

From  the  shape  of  the  curve  we  observe  that  the  rate 
of  variation  of  the  speed  is  not  uniform  and  that  for  small 
currents  the  speed  changes  more  rapidly  than  for  larger 
currents.  The  slope  of  the  curve  is  measured  by  the 
slope  of  its  tangent.  The  slope  of  the  curve  at  D  is  meas- 
ured by  the  slope  of  the  tangent  LD. 

The  tracing  of  the  curves  exemplifies  the  reversibility  of  all 
mathematical  processes.  It  is  just  as  easy  to  prepare  the 
data  for  a  table  from  a  curve  as  it  is  to  trace  a  curve  from 
the  data  of  the  table.  This  law  of  duality  has  been  illustrated 
in  every  pair  of  operations  in  the  preceding  chapters  and  mill 
recur  in  all  the  chapters  to  follow. 


Ex.  1.  Construct  the  magnetization  curve  for  a  machine 
from  the  data  in  Table  XII.  The  curve  is  shown  in  Fig.  109. 
The  extrapolation  of  the  curve  is  shown  by  the  dotted  line  and  is 

TABLE  XII.     MAGNETIZATION   CURVE   FOR  A  MACHINE 


Flux  per  pole  * 

1X106 

2X106 

3X106 

3.5  X106 

4X106 

Field  current  / 

13.18 

24.77 

38.63 

47.72 

59.09 

THE  USE  OF  SQUARED  PAPER 


285 


constructed  upon  the  assumption  that  when  there  is  no  field 
current  there  will  be  no  flux.  This  is  not  strictly  true  unless 
there  is  no  residual  magnetism. 

What  current  will  be  necessary  in  order  to  produce  a  flux 
=2.5  XlO6? 

What  flux  will  be  produced  by  a  current  of  30  amperes? 


4000000 


3500000 


8000000 


P  2500000 


S  2000000 

6 


1500000 


1000000 


500000 


/ 

/ 

/ 

/ 

r 

/ 
/ 

/ 

0  10  20  30  40  50  GO 

Current  in  Amperes 

I 

Fig.  109. — Magnetization  Curve  for  a  Machine. 


What  flux  will  be  produced  by  a  current  of  35  amperes?  In 
this  figure  as  well  as  for  subsequent  figures,  the  one-tenth  fines 
have  been  omitted  for  clearness. 

Ex.  2.  Construct  a  curve  from  the  data  in  Table  XIII.  This 
curve  is  shown  in  Fig.  110  in  which  the  three  points  A,  B,  and 
C  appear  to  be  on  a  straight  line.  Construct  the  figure  so  that 
the  flux  is  read  on  the  vertical  axis. 


286 


PRACTICAL  MATHEMATICS 
TABLE  XIII 


Point. 

Flux  3>. 

Ampere  Turns,  IN. 

A 

1,600,000  lines 

12,070 

B 

1,700,000     " 

13,207 

C 

1,800,000     " 

14,105 

14000 

D 

rC 

/ 

A 

13000 
10000 

¥ 

A 

E 

% 

Turns 

mpere 

< 

4000 

2000 

n 

500,000 


1,000,000 
Flux  in  Lines 


1,500,000 


2,000,000 


Fig.  110. — The  Normal  Position  of  Axes. 

4.  Transformation  of  the  Axes.  After  the  plotting 
of  Table  XIII  it  will  be  observed  that  the  curve  occupies 
a  very  small  part  of  the  paper.  The  value  of  the  curve 
would  be  enhanced  if  larger  scales  were  chosen.  To  increase 
the  scale  would  necessitate  a  larger  sheet  of  paper  if  we 
wished  to  preserve  the  axes  and  origin  with  the  curve. 
We  can  dispense  with  the  axes  since  we  are  concerned  with 
that  area  of  the  paper  only  which  is  occupied  by  the  curve. 


THE  USE  OF  SQUARED  PAPER 


287 


Consider  the  quadrangle  ADCE  enclosing  and  bounding 
the  curve  which  extends  horizontally  from  1.6  X106  to 
1.8  X106  and  vertically  from  12070  to  14105. 

We  shall  magnify  this  area  by  considering  the  point 
(12000,  1.6  X106)  moved  down  to  the  usual  position  of  the 
origin  at  the  lower  left-hand  marginal  intersection  as  shown 
in  Fig.  111.  Locate  the  point  1.8 X106  at  the  right- 
hand  limit  of  the  horizontal  axis.  The  two  points  1.6  X106 
and  1.8X106  will  be  separated  by  100  divisions.  In  like 
manner  locate  14105  near  the  upper  limit  of  the  vertical 
axis. 

Replot  the  curve  to  the  new  scale.  The  gain  in  size  is 
compensated  by  the  magnification  of  errors  incidental  to 
the  lack  of  abundant  data.  The  slope  of  the  curve  becomes 
better  defined  with  the  magnification. 

This  operation  is  called  the  transformation  of  the  coor- 
dinate axes.  It  is  virtually  equivalent  to  shifting  the  origin 
beyond  the  limits  of  the  paper.  The  scales  may  be  entered 
on  any  convenient  horizontal  or  vertical  bounding  line. 
The  work  is  simplified  by  plotting  megalines  instead  of  lines. 

Ex.  3.    Use  the  principle  of  paragraph  4  in  plotting  Table  XIV. 
TABLE  XIV 


Flux  <*>. 

Ampere  Turns,  IN. 

5000 

2500 

6500 

3175 

8250 
I 

5500 

Ex.  4.  Construct  curves  between  B  and  H  expressed  in 
gausses,  for  sheet  steel,  wrought  iron,  soft  cast  steel,  and  cast 
iron.  Plot  these  upon  the  same  sheet  of  paper,  using  the  data 
from  Table  XV. 

These  curves  are  called  magnetization  curves  because  they 
express  the  relation  between  the  flux  and  the  magnetizing  force. 


14000 


13500 


£13030 


12500 


12000 


A 

D 

/ 

/      1 
'         i 

i 

i 

/ 

i 

i 
i 

/ 

i 

Q 

| 

e! 

1.6 


1.65  1.7  1.75 

Flux  in  Millions  of  Lines 


1.8 


Pig.  111- 

-The  Transformation  of  the  Coordinate 

Axes. 

J^Ss: 

*T£6t 

16000 
15000 
14000 

12000 

G 

— 

^op 

n* 

vva<- 

- 

h/    S: 

l 

1 
1 

| 

1100    H 

1 

1 

100l>0 

If  40? 

sou    o 

/  < 
-     '  o 

700/    u> 

/     > 

ooo-b 

cf     * 

40O—| 
300     J 

1 

"cast" 

^bo^T 

8000 
6000 
4000 
2000 

c 

A 

K 

IjO 

K 

|E 

Q*,W 


20       30 


■10 


80 


50       l>0        70 
H    IN   GAUSSES 

Fig.  112. — Magnetization  Curves 


90 


ILK)      110      120 


288 


THE  USE  OF  SQUARED  PAPER 


289 


The  same  curves  may  be  used  to  express  the  relation  between 
the  flux  density  per  square  inch  read  vertically  and  the  ampere- 
turns  per  inch  of  length  read  horizontally  or  the  ampere-turns 
per  centimeter  of  length  read  horizontally.  Supplement  Fig. 
112  by  adding  the  three  scales  referred  to  above.  Add  the 
curve  for  sheet  steel. 

TABLE  XV.     VALUES  OF  B  AND  H  IN  GAUSSES 


H 

B 

Sheet  Steel. 

Cast  Steel.       ■  Wrought  Iron.  1      Cast  Iron. 

3,000 
3,500 
4,000 
4,500 

1.3 

2.8 

2.0 

5.0 
6.5 

3.4 

8.5 

11.0 

5,000 
5,500 
6,000 
6,500 
7,000 
7,500 
8,000 
8,500 

14.5 

18.5 

2.3 
2.4 

4.5 

3.5 

24.0 
30.0 

38.5 

49.0 

5.8 

60.0 

74.1 

10,000 
10,500 
11,000 
12,000 
13,000 
14,000 
15,000 
15,500 
16,000 
16,500 
17,000 
17,500 
18,000 

3.9 
4.1 

7.5 

124 

6.0 

6.5 

7.9 

10.0 

15.0 

25.0 

35.0 

49.0 

69.0 

93.0 

120 

9.0 
11.5 
Id 

21.5 
32.0 

5.0 

6.0 

9.0 

15.5 

27.0 
37.5 
52.5 
70.0 
92.0 

49.0 
60.0 
74.0 
93.0 
115 

Ex.  5.  Prepare  table  XVI  as  described  below,  using  the  values 
determined  from  the  four  curves  of  Ex.  4.  Divide  a  page  hori- 
zontally into  seven  columns  and  allow  space  under  each  column 
heading  for  13  entries.  The  first  column  will  be  headed  H  and 
under  it  make  12  entries  varying  in  steps  of  10  from    10  to  120 


290  PRACTICAL  MATHEMATICS 

inclusive.  Make  the  corresponding  entries  under  the  following 
column  headings:  Ampere-turns  per  centimeter  length,  Ampere- 
turns  per  inch  length;  under  double  columns  headed  respectively 
cast  iron,  cast  steel,  wrought  iron  and  sheet  steel,  make  an  entry 
for  B  in  kilo-gausses  and  also  an  entry  in  kilo-maxwells  per  square 
inch. 

5.  Plotting  of  a  Third  Variable.     In  the  discussion  of 
permeability  in  Chapter  IX,  page  185,  it  was  shown  that 

B 


*  =  H> 

whic 

h  may  be  written 

(i) 

[L      B 

1    #' 

Locate  any  point  such  as  F  on  one  of  the  curves.  Draw 
the  ordinate  FE.  Then  FE  =  the  B  of  the  point  (F)  and 
AE  =  the  H  of  the  point  (F).  Join  A  with  F.  Lay  off 
AL  =  \  unit,  draw  CL,  then  AF  intersects  CL  at  C. 


(2)  ~lT  =  Tn similar  Rt  As 


CL 

al" 

FE 
~AE 

CL 

B 

1 

H 

(3)      ...     r_  =±i  subs,  for  FE  and  AE  in  (2) 


(4)      .*.     CL=  [l from  (1)  and  (3)  and  =  fa/  ax 

For  every  point  on  the  curve  there  is  a  definite  point 
on. CL.  The  latter  point  is  the  \x  value  of  the  former  because 
it  expresses  the  ratio  of  the  B  to  the  H  of  the  former. 

The  values  of  y.  may  be  too  close  together  on  CL  for 
close  estimation  but  this  is  avoided  by  reading  [x  on  any 
other  line  such  as  KD  which  is  parallel  to  CL. 

Triangles  DKA  and  CLA  are  similar  right  triangles 
and  therefore  DK  =  10  (jl.  In  other  words  DK  is  propor- 
tional to  (a.     For  the  point  F,  B  =  15000  and  #  =  32  and 


THE  USE  OF  SQUARED  PAPER       291 

therefore  ^  =  468.  Therefore  either  C  or  D  may  be  marked 
468.  By  proceeding  in  like  manner  to  join  other  points 
on  the  curve  with  A  the  intersection  on  DK  will  give  the 
corresponding  [l  values  of  the  respective  points.  If  the 
points  are  carefully  selected  the  values  of  [k  may  be 
expressed  in  multiples  of  100. 

Ex.  6.  Calibrate  the  line  DK  in  gradations  of  100  by  using 
other  points  on  the  cast  steel  curve,  show  that  the  calibrations  of 
DK  apply  to  all  magnetization  curves. 

Ex.  7.  Logarithmic  Curve.  Consult  a  table  of  logarithms 
(three  places  will  suffice).  On  the  horizontal  axis  plot  numbers 
from  1  to  15.    On  the  vertical  axis  plot  the  corresponding  logarithms. 

On  the  same  sheet  plot  a  second  curve  for  Napierian  logs. 

Show  how  additional  or  supplementary  scales  may  enable 
the  curves  to  be  used  for  numbers  ranging  from  10  to  150,  or 
for  any  multiple  of  1  to  15. 

In  the  case  of  common  logarithms  the  vertical  scale  should 
read  mantissas  and  the  corresponding  characteristic  should  be 
added  mentally. 

All  logarithm  curves  pass  through  the  point  1  on  the  horizontal 
axis.     Why? 

In  Fig.  113  1LR  is  the  curve  of  common  logarithms  and 
UP  is  the  curve  of  Napierian  logarithms.  The  vertical  scale 
for  these  curves  is  the  one  closest  to  the  axis  and  will  be  called 
the  axal  or  first  scale.  A  second  scale  appears  further  to  the  left 
of  the  vertical  axis  and  applies  to  the  curve  1KQ.  The  latter 
is  a  magnified  curve  of  common  logarithms  and  has  been  plotted 
to  the  outside  scale  for  closer  reading.  1KQ  represents  the  common 
logarithms  of  squares  of  numbers  when  the  axal  scale  is  used. 
Why? 

The  Napierian  curve  may  be  obtained  graphically  from  the 
curves  1LR  and  1KQ. 

The  difference  between  the  corresponding  ordinates  of  UP 
and  \KQ  equals  .303  times  the  corresponding  ordinate  of  1LR. 
Two  of  these  differences  are  shown  as  d  and  c.  Construct  a 
right  triangle  with  a  base  line  equal  to  2.303  units  and  an  alti- 
tude ye.  Draw  yc  parallel  to  ye  within  the  triangle  at  one  unit's 
distance  from  the  vertex. 

We  then  obtain 

Vc     Jh 
2.303     1 ' 


292 


PRACTICAL  MATHEMATICS 


r-f 

1.2       2.4 
1.1       2.2 

c 

rA^ 

.9       1.8 
.8       1.6 

\ 

.7    mU 

a 

si 

too 

O 

M 
.5       1.0 

L 

r-AR 

.4         .8 
.3        .6 
.2        .4 
.1        .2 

0         0 

Cii 

d 

r 

% 

f/tt, 

, 

i 

1    N 

l_ 

c 

0      1      2       3      4       5      6      7      8      9     10     11     12    13     14    15 
Numbers 

Fig.  113. — Logarithmic  Curves, 


THE  USE  OF  SQUARED  PAPER  293 

and  therefore  v/f=2.303?/c.  This  graphic  multiplication  method 
may  be  applied  to  obtain  the  Napierian  curve  UP  from  the 
common  curve  1LR.  This  is  illustrated  as  follows.  Extend  a 
line  joining  L  with  9  until  it  intersects  the  ordinate  passing  through 
11.303.  Carry  the  intersection  horizontally  across  to  intersect 
LA''  and  thereby  locate  J  on  UP.  In  the  right  triangle  LN  and 
JN  will  be  the  measure  of  yc  and  ye  respectively. 

Ex.  8.  Periodic  Curves.  Consult  a  natural  sine  and  cosine 
table  (three  places  will  suffice). 

Construct  a  table  XVII,  showing  the  range  of  numeric 
variation  and  the  algebraic  signs  for  the  eight  trigonometric  functions 
with  regard  to  four  quadrants. 

What  angles  correspond  to  the  greatest  positive  and  greatest 
negative  values  of  their  sines?  What  are  the  angles  whose  sines 
are  zero? 

What  angles  correspond  to  the  greatest  positive  and  greatest 
negative  values  of  their  cosines?     What  is  cos  -1(0)? 

Turn  a  sheet  of  cross-section  paper  so  that  its  greatest  length 
is  horizontal  and  draw  the  horizontal  axis  through  the  center  of 
the  paper.  Beginning  at  the  left  end  of  the  axis  locate  points 
on  it  corresponding  to  every  15°  of  angle,  ranging  from  0°  to 
360°.  Through  these  points  of  division  draw  light  vertical  lines. 
Sin(;e 'sines  and  cosines  of  angles  lie  between  (+1)  and  (  —  1), 
then  20  divisions  in  the  vertical  direction  will  correspond  to  1.000 
in  the  tables.  Above  or  below  the  15°  points  of  division  depending 
upon  the  algebraic  sign,  lay  off  the  corresponding  values  of  the 
sines  of  the  respective  angles. 

For  values  beyond  90°  apply  the  definitions  of  sine  and  cosine 
with  due  regard  to  algebraic  signs. 

The  functions  of  angles  of  the  second  quadrant  have  numeric 
values  corresponding  to  the  like  named  functions  of  their  supple- 
ments. 

To  obtain  the  values  of  functions  for  angles  of  the  third  quad- 
rant subtract  180°  from  the  angle. 

To  obtain  the  values  of  functions  for  angles  of  the  fourth  quad- 
rant subtract  the  angle  from  360°. 

Through  all  the  plotted  points  pass  a  smooth  curve  which 
takes  the  name  sine  curve  according  to  the  name  of  the  plotted 
function. 

In  a  like  manner  construct  a  cosine  curve.  Use  the  same  axis 
and  the  same  scale  which  was  used  for  the  sine  curve.    See  Fig.  114. 

What  are  the  features  of  similarity  and  dissimilarity  in  the 
sine  and  cosine  curves? 


294 


PRACTICAL  MATHEMATICS 


0 


saujg 
sauisoQ 


THE  USE  OF  SQUARED  PAPER  295 

Show  from  the  curves  that  the  sine  of  an  angle  of  the  first 
quadrant  equals  the  cosine  of  its  complement. 

That  portion  of  the  curves  lying  between  the  zero  ordinates 
of  the  curves  is  called  an  arc  or  arch  or  loop.  Two  consecutive 
arches  constitute  a  complete  cycle. 

Where  two  curves  appear  on  one  sheet  of  paper  they  should 
be  marked  in  some  manner  so  as  to  be  identified  easily.  Thus 
in  Fig.  114  the  sine  curve  is  marked  (a)  and  the  cosine  curve 
is  marked  (b). 

The  curves  may  be  extended  indefinitely  to  the  right  and  to 
the  left  of  the  zero  division.  It  will  be  seen  that  the  curves  repeat 
in  both  directions  with  a  definite  rhythm  or  in  precisely  equal 
intervals  on  the  axes  and  hence  are  said  to  be  periodic. 

6.  Plotting  of  Formulas.  We  have  seen  that  a  series 
of  experimental  results  or  a  calculated  logarithmic  or  trig- 
onometric table  may  be  plotted  and  a  curve  drawn  so  as 
to  pass  as  evenly  as  possible  through  the  points.  It  is 
not  possible  to  draw  a  smooth  curve  through  all  the  plotted 
points  of  experimental  data.  About  an  even  number  of 
points  should  be  distributed  on  both  sides  of  the  curve. 
Misplaced  points  are  due  to  the  errors  of  observation  and 
experimental  defects  and  therefore  the  resulting  curve 
should  approximate  to  the  average  of  these  deviations. 
Data  which  when  plotted  show  large  deviations  should  be 
discarded. 

Any  formula  may  have  two  of  its  letters  regarded  as 
variables.  The  remaining  letters  will  be  considered  fixed  or 
constant  while  the  variables  will  change  through  a  range 
of  coordinated  or  associated  values.  If  we  substitute 
assumed  values  for  one  variable  it  is  called  an  independent, 
then  the  other  variable  is  called  a  dependent  and  the  latter 
may  he  solved  for  definite  values  according  to  the  relation 
of  the  variables  in  the  formula.  These  pairs  of  values 
may .  be  tabulated.  The  tables  may  be  used  to  plot  a 
curve. 

In  this  way  the  curve  becomes  the  pictorial  represen- 
tation of  the  formula. 


296  PRACTICAL  MATHEMATICS 

A  definite  curve  corresponds  to  every  formula  and  conversely 
a  definite  formula  corresponds  to  every  curve. 

In  the  paragraphs  which  follow  methods  are  illustrated 
for  deriving  the  equation,  i.e.,  formula  corresponding  to 
a  curve. 

W 
7.  The  Graphic  Representation  of  Ohm's  Law.    /=«•• 

The  formula  of  Ohm's  Law  contains  three  quantities  J, 
E,  and  R.     The  values  of  I  will  depend  upon  the  value 

E 

of  the  ratio  — .     If  we  consider  a  circuit  having  a  constant 
R 

resistance,  R  =  5  ohms,  then  as  the  impressed  voltage  E 
is  made  to  vary,  there  will  be  a  corresponding  variation  in 
I.  For  every  value  assigned  to  E  there  will  be  a  correspond- 
ing dependent  value  of  I.  If  E  is  made  to  vary  from  0  to 
10  volts  in  gradations  of  1  volt  then  by  substituting  these 
values  and  also  the  value  of  R  =  5  we  obtain  a  corresponding 
range  of  values  for  7. 

The  associated  pairs  of  values  of  E  and  I  are  arranged 
and  entered  in  order  in  Table  XVIII  as  shown.  It  is  advis- 
able to  fill  in  the  entire  E  column  first  by  beginning  with 
the  least  value,  i.e.,  0,  and  ending  with  the  highest  assigned 
value  10.  As  each  value  of  I  is  calculated  make  the  entry 
in  the  I  column  opposite  the  corresponding  value  of  E. 

E     0 
Suppose    E  =  0   then   I  =  ^  =  ~=0.     Under    the    I  column 

R     o 

enter  0  opposite  0  in  the  E  column.     When  E=\  then 

I  =  \  =  .2.     Under   the   I   column   enter    .2   opposite    1    in 

the  E  column.     Finally  when  E  =  10  then  I  =  V  =  2.     Under 

the  I  column  enter  2  opposite  10  in  the  E  column.     From 

the  data  so  obtained  we  proceed  to  plot  the  corresponding 

curve  in  the  usual  manner. 

Plot  these  values  labeling  the  longer  axis  E  and  the 

shorter  axis  I.     The  resulting  graph  or  curve  is  a  straight 

line  but  a  straight  line  is  also  a  curve  in  the  graphic  sense. 


THE  USE  OF  SQUARED  PAPER 
TABLE  XVIII.     R  =  5tt. 


297 


E 

I 

E 

/ 

0 

0 

6 

1.2 

1 

.2 

7 

1.4 

2 

.4 

8 

1.6 

3 

.6 

9 

1.8 

4 

.8 

10 

2.0 

5 

1.0 

The  curve  passes  through  the  origin.  This  condition 
will  occur  whenever  the  values  of  both  variables  are  zero 
simultaneously. 

The  curve  is  said  to  be  the  locus  of  the  formula,  i.e., 
the  position  collectively  of  every  point  whose  pairs  of 
values  validate  the  formula  upon  substitution. 

The  pair  of  values  corresponding  to  any  piont  are  the 
respective  distances  measured  parallel  to  the  axes. 

They  are  spoken  of  collectively  as  the  coordinates  of 
a  point. 

For  purposes  of  distinction  the  first  mentioned  distance 
"  the  abscissa  "  is  measured  always  from  the  vertical  axis. 

The  second  mentioned  distance  "  the  ordinate "  is 
measured  always  from  the  horizontal  axis. 

Points  are  designated  by  enclosing  their  abscissa  and 
ordinate  values  in  a  parenthesis. 

P  =  (l,  0.2)  means  P  is  the  point  corresponding  to  an 
abscissa  1  and  an  ordinate  0.2.  A  comma  separates  the 
two  enclosed  values  and  a  correspondence  symbol  (  =  )  is 
placed  between  a  letter  and  the  corresponding  parenthesis. 

P  =  (8,  1.6)  means  locate  a  point  8  units  distance  along 
the  horizontal  axal  direction  and  1.6  unit  in  the  vertical 
axal  direction.  This  is  a  convention  that  must  be  observed 
for  the  above  notation,  although  the  point  might  be  located 
equally  well,  by  proceeding  first  in  the  vertical  direction 
and  then  in  the  horizontal  direction. 


298 


PEACTICAL  MATHEMATICS 


The  carrying  capacity  of  insulated  copper  wires  for  interior 
wiring  according  to  the  National  Electric  Code  is  given  in 
Table  XIX. 

TABLE  XIX.  CARRYING  CAPACITY  OF  INSULATED 
COPPER  WIRES  FOR  INTERIOR  WIRING,  NATIONAL 
ELECTRICAL  CODE 


A 

B 

C 

D 

E 

F 

G 

Rubber 

Weather 

Rubber 

Weather 

B.  &S. 

Circular 

Covered 

Proof 

Circular 

Covered 

Proof 

Co. 

Mils. 

Wires, 

Wires, 

Mils. 

Wires, 

Wires, 

Amperes. 

Amperes. 

Amperes. 

Amperes. 

18 

1,624 

3 

5 

200,000 

200 

300 

16 

2,583 

6 

8 

300,000 

270 

400 

14 

4,107 

12 

16 

400,000 

330 

500 

12 

6,530 

17 

23 

500,000 

390 

590 

10 

10,380 

24 

32 

600,000 

450 

680 

8 

16,510 

33 

46 

700,000 

500 

760 

6 

26,250 

46 

65 

800,000 

550 

840 

5 

33,100 

54 

77 

900,000 

600 

920 

4 

41,740 

65 

92 

1,000,000 

650 

1000 

3 

52,630 

76 

110 

1,100,000 

690 

1080 

2 

66,370 

90 

131 

1,200,000 

730 

1150 

1 

83,690 

107 

156 

1,300,000 

770 

1220 

0 

105,500 

127 

185 

1,400,000 

810 

1290 

00 

133,100 

150 

220 

1,500,000 

.  850 

1360 

000 

167,800 

177 

262 

1,600,000 

890 

1430 

0000 

211,600 

210 

312 

1,700,000 
1,800,000 
1,900,000 
2,000,000 

930 

970 

1010 

1050 

1490 
1550 
1610 
1670 

Ex.  9.  Label  the  horizontal  axis  for  the  A  and  B  scales  and 
the  vertical  axis  for  the  C  and  D  scales.  Plot  one  curve  for  values 
of  A  and  C  and  another  curve  for  values  of  A  and  D. 

Ex.  10.  Label  the  horizontal  axis  E  and  the  vertical  axis 
F  and  G.  Plot  one  curve  for  values  of  E  and  F  and  another 
curve  for  values  of  E  and  G. 

Ex.  11.  Plot  the  following  table  XX  for  temperature  coefficient 
of  copper  at  different  initial  temperatures  in  Centigrade.  Sup- 
plement the  Centigrade  scale  with  a  scale  for  reading  Fahrenheit. 


THE  USE  OF  SQUARED  PAPER 


299 


TABLE  XX.     TEMPERATURE   COEFFICIENTS 


Initial 

Temperature 

Initial 

Temperature 

Temperature 

Coefficient  in  Per 

Temperature, 

Coefficient  in  Per 

Centigrade. 

Cent  per  Degree 
Centigrade. 

Centigrade. 

Cent  per  Degree 
Centigrade. 

0 

.420 

26 

.379 

1 

.418 

27 

.377 

2 

.417 

28 

.376 

3 

.415 

29 

.374 

4 

.413 

30 

.373 

5 

.411 

31 

.372 

6 

.410 

32 

.370 

7 

.408 

33 

.369 

8 

.406 

34 

.368 

9 

.405 

35 

.366 

10 

.403 

36 

.365 

11 

.402 

37 

.364 

12 

.400 

38 

.362 

13 

.398 

39 

.361 

14 

.397 

40 

.360 

15 

.395 

41 

.358 

16 

.394 

42 

.357 

17 

.392 

43 

.356 

18 

.391 

44 

.355 

19 

.389 

45 

.353 

20 

.388 

46 

.352 

21 

.386 

47 

.351 

22 

.385 

48 

.350 

23 

.383 

49 

.348 

24 

.382 

50 

.347 

25 

.380 

CHAPTER  XV 
LINEAR  GRAPHS 

1.  In  the  graphic  work  which  follows  prepare  the  tables 
completely  before  proceeding  to  plot.  Begin  by  sub- 
stituting a  zero  value  or  the  least  assigned  value  for  one 
of  the  variables.  Label  curves  to  correspond  to  the  tables. 
Designate  the  axes  in  full  and  with  symbolic  abbreviations. 
Where  the  subject  matter  bears  a  title  enter  it  upon  the 
cross-section  paper.  Tables  may  be  entered  on  the  cross- 
section  paper  but  it  is  preferable  to  enter  them  with  the 
numeric  calculations  upon  a  sheet  of  plain  paper  which 
is  inserted  in  the  work  book  so  as  to  face  the  graph. 

The  cross-section  paper  upon  which  the  graph  has  been 

drawn  is  called  a  plate  and  is  numbered  in  Roman  notation 

and  bears  a  cross  reference  to  the  page  of  tables,  calculations 

and  other  descriptive  matter. 

E 
Ex.  1.    Prepare  a  table  of  values  and  plot  I  =— ,  wherein  R 

is  constantly  equal  to  1  ohm  and  E  varies  from  0  to  12  volts. 
This  curve  is  shown  as  ACE  in  Fig.  115.  The  (a)  scales  apply 
to  it. 

Ex.  2.  Variation  of  Coulombs  with  Time.  Prepare  a  table 
of  values  and  plot  Q=It,  where  /  is  constantly  equal  to  10  amp. 
and  t  varies  from  0  to  60. 

This  curve  is  shown  as  LMN  in  Fig.  115.  The  (6)  scales 
apply  to  it. 

Ex.3.  Variation  Metallic  Deposition  with  Current.  Prepare 
a  table  of  values  and  plot 

W 
I  =—-;  K2  is  comstantly  =0.0003386, 
K2t 

t  is  constantly  =  1  hr.  =60  min.  =3600  sec. 

/  varies  from  2.5  to  10. 

300 


LINEAR  GRAPHS 


301 


This  curve  is  shown  as  QS  in  Fig.  115.     The   (c)   scales  apply 
to  it. 


10       12       14 
«.«)  VOLTS 


15      20       25       30      35       40      45        50 
(b)  TIME 


4        5         6 
(C)  WEIGHT 


10       11 


1.2      1.5      1.8     2.1     2.4      2.7 
(d)  AMPERES 


Fig.  115. — The  Graphs  of  Simple  Equations. 

Ex.  4.     Variation  Liberation    Gas  with  Current. 

a  table  of  values  and  plot 


Prepare 


/  =77~;  Ki  is  constantly  =0.1733, 
Kit 

t  is  constantly  =  100  sec. 
I  varies  from  0.3  to  2.7. 

This  curve  is  shown   as  GHJ  in  Fig.  115.     The  (d)    scales 
apply  to  it. 


302  PRACTICAL  MATHEMATICS 

Ex.  5.    On  one  sheet  of  cross-section  paper  plot  three  graphs 

E 
of  /=•£.     These  may   be   obtained  by  preparing  three   distinct 

tables  regarding  E  and  /  as  the  variables  but  by  assigning  a 
definite  value  to  R  for  each  table  as  follows:  (a)  when  #=.112, 
(b)  when  R=10ti,  (c)  when  ft  =  100&  Label  the  graphs  (a), 
(6)  and  (c)  to  correspond  to  the  tables.  Plot  /  vertically  and 
E  horizontally.  From  the  preceding  plots  of  Ohm's  Law  it  was 
observed  that  in  each  case  the  graph  is  a  straight  line.  The 
three  graphs  of  Ex.  5  will  be  straight  lines.  A  straight  line  is 
completely  determined  by  two  of  its  points  and  therefore  the 
above  tables  should  contain  only  two  pairs  of  entries.  Since 
one  of  the  pairs  of  values  indicates  that  the  line  passes  through 
the  origin,  it  is  advisable  for  accuracy  that  the  other  pair  of  values 
should  locate  a  point  at  considerable  distance  from  the  origin. 

2.  The  slope  of  a  straight  line  is  the  numeric  value 
of  the  tangent  of  its  angle  of  inclination  to  the  hori- 
zontal when  the  scales  of  both  axes  are  equal.  Whenever 
the  scales  of  the  two  axes  are  unequal  the  slope  is  the 
ratio  of  the  number  of  units  in  the  perpendicular  to  the 
number  of  units  in  the  corresponding  projection  of  the 
line. 

In  Fig.  115  the  slope  of 

,_,    EF    8     \      ,,      .  ,  _,__    NP     300     in 

AE=Qp  =  g  =  l,    the  slope  of  Li\T  =  — =—  =  10. 

Ex.  6.    Determine  the  slope  of  QS  and  GJ  in  Fig.  115. 

Ex.  7.  Determine  the  slopes  of  graphs  a,  b  and  c  in  Ex.  5. 
How  does  the  value  of  R  affect  the  slope?  How  does  the  value 
of  the  reciprocal  of  R  affect  the  slope?  What  is  the  relation  of 
reciprocal  of  R  to  El 

3.  It  is  customary  to  plot  the  dependent  variable 
vertically  and  the  independent  variable  horizontally.  The 
slope  will  be  numerically  the  same  as  the  coefficient  of 
the  independent  variable.  In  any  equation  either  variable 
may  be  made   the  dependent   variable.    The   formula  or 


LINEAR  GRAPHS  303 

equation  must  be  transformed  so  that  the  dependent  varia- 
ble has  a  coefficient  and  exponent  equal  to  unity. 

Following  the  convention  of  signs  established  for  trig- 
onometry distances  measured  to  the  right  and  above  the 
origin  are  positive,  and  on  the  contrary  distances  measured 
to  the  left  and  below  the  origin  are  negative. 

Ex.8.  Centigrade-Fahrenheit  Conversion.  PlotC  =  f(F-32). 
Transform  and  plot  F  as  the  dependent  variable.  Draw  the  axes 
through  the  center  of  the  paper.  The  range  of  C  is  to  be  from 
-300°  to  +300°.  This  graph  is  illustrated  as  (a)  in  Fig.  116. 
The  line  is  not  extended  beyond  -273°  C.  Why?  Measure  the 
slope  of  the  graph  and  compare  it  with  the  coefficient  of  the  inde- 
pendent variable.  What  is  the  value  of  the  intercepted  distance 
on  the  vertical  axis  between  the  origin  and  the  graph? 

Ex.  9.  Copper  Temperature  Coefficient.  Plot  Rt  = 
#o(l+.0O4277).  Ro  =9.59  ohms  which  is  the  mil  foot  resistance 
of  annealed  copper  wire  at  0°  C.  The  values  of  T  are  in  centi- 
grade and  range  from  —20°  to  +50°.  This  graph  is  shown  as  (6) 
in  Fig.  116.  Measure  the  slope  of  the  graph  and  show  that  it 
is  numerically  the  same  as  .0042i?o.  By  means  of  the  Centigrade- 
Fahrenheit  conversion  scales  show  that  .0024  is  the  coefficient  of 
T  when  expressed  in  Fahrenheit  degrees.  The  value  of  R0  is  read 
off  at  the  point  where  the  graph  cuts  the  vertical  axis.  Plot  Ex. 
9  on  the  same  plate  as  Ex.  8,  using  Rt  &s  the  dependent  variable. 

Ex.  10.    Variation  of  E.M.F.  and  Resistance. 

E 
Plot  J  =  B^r~-  I  constantly  =10  amps., 

r  constantly  =  2  ohms, 
R  varies  from  0  to  10  ohms. 

4.  The  plots  of  examples  (1)  .  .  .  (10)  are  characterized 
by  graphs  which  are  straight  lines.  The  slopes  of  the 
straight  graphs  as  well  as  their  intersections  with  the  axes 
of  reference  is  very  easily  predetermined  before  the  formula 
is  plotted. 

The  formulas  that  we  have  dealt  with  are  of  the  first 
degree   and   are   represented   by   straight   line   graphs.    A 


304 


PRACTICAL  MATHEMATICS 


formula  is  said  to  be  linear  when  it  is  an  equation  of  the 
first  degree.  Why  are  the  formulas  of  Ex.  (1)  .  .  .  (10) 
of  the  first  degree? 


Fig.  116. — (a)  Centigrade-Fahrenheit  Conversion. 
(6)  Copper  Temperature  Coefficient. 

The  formulas  (1)  .  .  .  (10)  may  be  regarded  as  forms 
of  one  simple  equation,  viz.: 


(i) 


y  =  a-\-bx. 


LINEAR  GRAPHS  305 

This  is  shown  by  assigning  to  x,  y,  a,  and  b  those  values 
which  will  identify  (1)  with  each  equation  in  the  group 

(1)  .  .  .  (10). 
To  show  that 

E 

(2)  7=pisa  form  of  y  =  a+bx, 

make  y  in  (1)=/  in  (2), 
.-«       x    <<    <<    E     .< 

"  ''■'■* : 

14       a    ' '    M    zero  in  (2)  (missing  term) 

E 

then  2/  =  a+for  becomes  I  =  —  and  the  latter  is  said  to  be 

a  special  form  of  the  linear  equation  y  =  a-\-bx. 
To  show  that 

(3)  Q  =  It  is  a  form  of  ?/  =  a+foc, 

make  ?/  in  (1)=Q  in  (3), 
"     x    "  •"      *     M 
"     &  "    M     7    '; 
"     a    "    tl    zero  in  (3)  (missing  term). 

To   show  #r=#0(l+0.00477)    is   a  form   of  y  =  a+bx, 
first  transform  it,  then  RT  =  R0-\-0.004ItoT. 

Then  show  that 

(4)  RT=Ro+0.00±RoT 
is  a  form  of  y  =  a-\-bx 

make  y  in  (l)=J?r         in  (4) 

i.       ^    m      M      rp 

"     b    "    "    0.004^o     " 
V     a    "    M    #0 

then  y  =  a-\-bx  becomes  Rt=Ro-\-0.004RoT  and  therefore 
#r  =  #o(l+0.004T)  is  said  to  be  a  special  form  of  (1). 


306  PRACTICAL  MATHEMATICS 

Show  how  the  formulas  in  Ex.  (3),  (4),  (8),  and  (10) 
are  interpreted  as  special  forms  of  the  linear  equation 
y  =  a+bx. 

6.  Degree  of  an  Equation.  We  have  noted  that  equations 
contain  two  kinds  of  quantities,  variable  elements  (variables) 
and  fixed  quantities  (constants).  The  latter  are  generally 
numeric  or  literal  factors  and  constitute  the  coefficients 
of  the  variables  in  their  respective  terms.  There  may  be 
one  isolated  constant  called  the  absolute  term. 

Variables  enter  equations  with  exponents  which  may  be 
integral  or  fractional,  numeric  or  literal,  positive  or  negative. 

The  degree  of  a  term  is  the  numeric  value  of  the  exponent 
of  the  variable  in  that  term.  When  several  variables  occur 
as  factors  in  a  term  then  the  degree  of  the  term  equals 
the  sum  of  the  exponent  of  the  variables. 

The  degree  of  an  equation  is  equal  to  the  degree  of  its 
highest  term. 

The  degree  of  an  equation  is  determined  only  after 
it  is  free  from  fractional  and  negative  exponents,  radicals 
and  denominators  with  variable  factors. 

Why  does  y  =  a-\-bx  represent  an  equation  of  the  first 
degree?  What  is  the  degree  of  the  group  of  equations 
which  are  special  forms  of  (1)?    Why? 

Ex.  11.    State  the  degree  of  the  following  equations: 
(5)  x=y2 


(6)  x=ay3 

(7)  x=c+dy 


(8)  x  = 


x  and  y  are  variables 


y 

(9)  pvn  =R p  and  v  are  variables 

(10)  W  =IE I  and  E  are  variables. 


LINEAR  GRAPHS 


307 


(11) 

'"S 

(12) 

I-    E 

R+r 

(13) 

5y=2+0.3x 

(14) 

y=7-5x 

(15) 

y=3-2.5x  J 

I  and  R  are  variables. 


x  and  y  are  variables. 


6.  In  the  preceding  examples  we  have  seen  that  every 
first  degree  equation  is  plotted  as  a  linear  graph,  i.e.,  a 
straight  line.  Two  questions  arise:  first,  is  every  equation 
of  the  first  degree  represented  by  a  linear  graph,  and  second, 
shall  we  interpret  every  linear  graph  as  an  equation  of  a 
straight  line?  We  shall  now  proceed  to  the  proof  of  these 
statements,  x  is  the  independent  (variable),  y  the  de- 
pendent (variable)  and  a  and  b  are  fixed  numeric  values 
(constants)  as  expressed  in  the  general  equation  of  the 
first  degree: 


(i) 


y  =  a+bx. 


Whatever  is  true  of  (1)  is  true  in  general  for  all  first 
degree  equations.  Assign  to  x,  any  six  values  designated 
by  xi,  Xo,  X2,  xs,  X4,  X5,  and  then  substitute  these  in  suc- 
cession in  (1),  and  obtain  the  corresponding  values  yi,  yo, 
2/2,  2/3,  2/4,  2/5,  from  (1)  as  shown  in  (a),  (b),  (c),  (d),  (e),  (/)• 
Letters  are  preferable  to  numbers  in  this  discussion. 


(a) 
(6) 
(c) 

(d) 

(e) 
(/) 


y!=a+bxi, 
yo  =  a+bxo, 
yi=a+bx2, 
y3  =  a+bx3, 

I/4=a+&X4, 

j/5  =  a+6x5, 


308 


PRACTICAL  MATHEMATICS 


These  pairs  of  values  are  tabulated  in  Table  XXI  and 
are  then  plotted  as  points  Pi,  Po,  P2,  P3,  P4,  and  P5  in 
Fig.  117. 


TABLE  XXL     COORDINATES  OF  POINTS 


Plotted  point 

Pi 

Po 

P2 

Pz 

Pa 

P5 

Values  of  x 

Xi 

Xq 

x2 

Xz 

Xi 

x5 

Values  of  y 

yi 

Vo 

yi 

2/3 

2/4 

2/5 

Construct  the  ordinates  at  each  point,  and  from  each 
point  and  between  the  adjacent  ordinates  draw  parallels 


Fig.  117. — A  Linear  Graph  Corresponds  to  Simple  Equation. 


to  the  XXf  axis.  These  construction  lines  are  represented 
by  dash  lines.  Join  each  pair  of  consecutive  points  by 
straight  lines:  P5P4,  P4P3,  P3P2,  P2P0,  P0P1.  In  this 
manner  we  have  formed  five  right  triangles  V,  IV,  III, 
II,  and  I.  The  pairs  of  sides  are  2/5  —  2/4  and  X5—X4,  2/4—2/3 
and  x±—xs,  2/3  —  2/2  and  X3—  X2,  2/2  —  2/0  and  £2— #o,  and 
2/0—2/1  and  xq— x\.    These  pairs  of  sides  have  equal  ratios 


LINEAR  GRAPHS  309 

which  may  be  obtained  by  subtracting  the  six  consecutive 
equations  in  the  group  (a),  (6),  (c),  (d),  (e),  (/)  above. 

(Q\    •    ys-y* =y*-y3  =V3-y2 =y2—yo =yo-yi  =b 
xs—x±    £4—23     £3  —  22     £2  — #o     xq=x\ 

These  equal  ratios  express  the  value  of  the  slopes  of 
P5P4,  P4P3,  P3P2,  P2P0,  and  P0Pi,  i.e.,  the  tangents  of 
5, 4,  3,  2,  and  1  are  equal  and  therefore  angles  5  =  4  =  3  =  2  =  1. 
Each  of  the  five  triangles  contain  a  right  angle.  Therefore 
the  sum  of  the  component  angles  at  each  of  the  points, 
P4,  P3,  P2,  and  P0  equals  180°.  In  other  words  P5P4P3 
is  a  straight  angle  and  therefore  P5P3  is  a  straight  line. 
By  a  like  authority  the  following  pairs  of  lines  form  straight 
angles,  P4P3  and  P3P2,  P3P2  and  P2Po,  and  P2Po  and  P0Pi. 
Therefore  P5,  P4,  P3,  P2,  Po,  Pi  all  lie  on  one  straight 
line  LN  and  since  these  points  were  plotted  from  any  assumed 
values  which  would  satisfy  (1)  as  shown  in  (a),  (6),  (c), 
(d),  (e),  (/),  then  all  points  which  satisfy  (1)  will  lie  on 
LN  or  LN  produced.  Conversely,  the  slope  measured  at 
every  point  on  LN  will  satisfy  equation  (1).     Therefore: 

Every  equation  of  the  first  degree  is  represented  by  a  straight 
line  and  every  straight  line  is  interpreted  as  an  equation  of 
the  first  degree. 

7.  Intercepts.  When  we  substitute  in  (1)  the  value 
for  x  =  0  then  (1)  reduces  to  y  =  a-\-bX0  =  a. 

This  means  that  the  line  cuts  the  Y  axis  at  a  distance 
a  from  the  origin.     This  distance  is  called  the  Y  intercept. 

On  the  other  hand,  if  the  equation  is  solved  for  x  and 
the  value  for  y  =  0  is  substituted  the  equation  reduces  to 

a 

This  means  that  the  line  cuts  the  X  axis  at  a  distance 

— -   from   the  origin.     This   is   a   negative   distance,   and 
0 


310  PRACTICAL  MATHEMATICS 

therefore  it  is  laid   off  to  the   left  of  the  origin.     It  is 
designated  as  the  X  intercept. 

8.  The  Slope  of  a  Linear  Equation.  The  line  LN 
(y  =  a+bx)  makes  an  angle  1  with  theX  axis.     The  slope  of 

LN=y_2=m 

X2  —  XQ 

but 

x2  —  Xq 

by  equation  (g). 

Therefore  the  slope  of  LN  =  b,  which  is  the  coefficient  of 
the  independent  variable  x  in  y  =  a-\-bx. 

In  equation  y  =  a-\-bx  we  have  by  recapitulation: 

a  is  the  intercepts  on  the  Y  axis, 

a 

— —  is  the  intercepts  on  the  X  axis, 
b 

b  is  the  slope  of  the  line. 

In  Fig.  117  a  line  TW  is  drawn  parallel  to  LN  so  as 
to  pass  through  0.  What  is  its  intercepts  on  the  Y  axis? 
Therefore  a  =  0. 

The  slope  of  TW  equals  the  slope  of  LN. 

Therefore  the  equation  of  a  line  passing  through  the 
origin  is  y  =  bx. 

Every  linear  equation  from  which  the  absolute  term 
(a),  i.e.,  constant  term,  is  missing  represents  a  line  through 
what  point? 

For  such  a  line  show  that  its  slope  = 

the  ordinate  of  a  point 


the  abscissa  of  the  same  point" 


LINEAR  GRAPHS  311 

Equation  (1)  may  be  transformed  into  (h)  in  which  the 
denominators  of  x  and  y  represent  the  intercepts  measured 
in  the  X  and  Y  directions  respectively. 

(h)  £+— ~L 

a      —a 


Observation.    Plotting   a   First   Degree  Equation.     Two 

points  are  sufficient  to  determine  a  straight  line.  Therefore 
before  attempting  the  plotting  of  formulas  of  the  first  degree 
determine  the  intercepts  of  the  graph.  Then  through  the  pair 
of  points  so  determined  draw  a  straight  line  which  corre- 
sponds to  the  formula. 

This  is  quickly  done  by  substituting  y  =  0  and  solving 
for  x,  then  substituting  x  —  0  and  solving  for  y. 

Should  x  and  y  both  equal  zero  simultaneously,  then  the 
line  passes  through  the  origin.  In  such  cases  an  additional 
pair  of  values  will  have  to  be  used  for  a  second  point.  There 
is  the  alternative,  however,  of  using  the  slope  at  the  origin. 

9.  Writing  the  Equation  for  a  Given  Straight  Line. 
I.  The  equation  of  a  straight  line  may  be  obtained 
by  substituting  in  (1)  or  (h),  the  value  of  b,  which  is  the 
slope  of  the  graph,  and  the  value  of  a,  which  is  the  inter- 
cepts on  the  Y  axis. 

II.  The  equation  of  a  straight  line  may  be  obtained 

by   substituting  in   (h)  the  values  of  a  and  — — ,  which  are 

o 

the  respective  intercepts  of  the  graph  on  the  Y  and  X  axes. 

III.  It  is  always  possible  to  measure  the  slope  of  a  line 
but  not  always  convenient  to  determine  the  intercepts.  In 
such  cases  make  use  of  (g).  Determine  6  the  slope  of  the 
line  and  the  coordinates  (xiyi)  for  any  point  on  the  line. 
Substitute  these  three  values  in  (g)  and  simplify. 


312  PEACTICAL  MATHEMATICS 

IV.  A  fourth  method  used  in  obtaining  the  equation  of 
a  straight  line  is  of  more  general  application  to  all  classes 
of  curves. 

We  wish  to  express  the  line  in  the  form  (1)  and  therefore 
it  is  necessary  to  determine  the  values  of  a  and  b.  Two 
points  on  the  curve  give  us  the  coordinates  Pi  =  (xiyi)  and 
p25s(afe#2).  Substitute  the  coordinates  in  (1)  and  obtain 
(a)  and  (c), 

(a)  yi^a+bxi, 

(c)  y2  =  a+bx2. 

From  the  equations  (a)  and  (c)  the  constants  a  and  b 
are  determined  by  one  of  the  methods  of  elimination  for 
the  solution  of  simultaneous  equations. 

10.  Graphic  Representation  of  Simultaneous  Equations. 
Two  linear  equations  when  plotted  give  two  corresponding 
straight  lines  which  by  geometry  intersect  in  a  single  point. 
This  commom  point  has  a  pair  of  coordinate  values  which 
when  substituted  in  both  given  equations  satisfies  or  validates 
them.  No  other  point  on  either  line  validates  both  equa- 
tions. The  coordinates  of  the  intersection  should  agree 
with  the  numeric  solution  obtained  by  elimination,  using 
algebraic  methods. 

Ex.  12.  Determine  the  solution  of  the  following  simultaneous 
equations: 

(a)  3z  +  y  =3, 

(6)  5z+2?/  =4. 

Determine  the  intercepts  for  (a)  and  enter  these  in  a  table 
labeled  (a)  and  then  determine  the  intercepts  for  (6)  and  enter 
these  in  a  table  labeled  (6).     Plot  directly  on  the  axes  from  the 


LINEAR  GRAPHS 


313 


tables.     Extend  the  linear  graphs  until  they  intersect.     Determine 
the  abscissa  and  ordinate  of  the  point  of  intersection  and  write 


V 

\  Va) 

(&W 

-1 

1                   1 

1                  3 

'     \ 

-1 

-2 

-9 

(2,-3) 

(a)3x  +  y 

=  3 

(&)5x  +  2l/  =  4 

-4. 

-R 

Fig.  118. — Graphic  Solution  of  Simultaneous  Linear  Equations. 


these  respective  values  for  x  and  y.    These  graphs  are  illustrated 
in  Fig.  118.     The  intersection  gives  x=2  and  y  =  —  3. 

An  enlargement  of  the  horizontal  scale  would  define  the  inter- 
section more  distinctly. 


314 


PEACTICAL  MATHEMATICS 


Ex.  13.    Determine  the  solution  of  the  following  simultaneous 
equations: 


(a) 
(b) 


5x—5y=25, 
7x+fy=  2. 


These  equations  are  illustrated  in  Fig.  119.    Upon  examination 
it  is  seen  that  the  intercepts  for  (6)  are  close  together  and  there- 


V 1 

i 

I 

5 

-1 

-2 

V) 

X) 

-3 

-4 

V2.-3) 

(a)  5x  —  5 
(6)7a;+4 

y=25 
y=-  2 

V 

Fig.  119. 


fore  a  third  point  should  be  plotted  for  (b)  to  give  a  more  accurate 
result. 

Prepare  the  tables  for  (a)  and  (b)  and  check  the  graphic  by 
the  algebraic  solution. 

The  first  step  is  to  simplify  (a)  by  dividing  it  by  5. 


LINEAR  GRAPHS 


315 


Ex.  14.  Determine 
the  solution  of  the 
following  simultaneous 
equations: 

(a)  4x—6y  =  8, 

(6)    Qx-y   =25. 

These  equations  are 
illustrated  in  Fig.  120. 
An  examination  of 
the  figure  shows  an 
extreme  case  in  which 

(b)  should  not  have 
been  drawn  through 
the  y  axis.  A  point 
midway  between  the 
two  plotted  points 
would  have  given  an 
equal  accuracy  with 
considerable  saving  of 
paper.  Prepare  the 
tables  accordingly  and 
check  the  work  by  an 
algebraic  solution. 


yi 

0 

(4 

44,1.6 

W 

-2 

/i 

1 

i 

\ 

1 

'-% 

y 

—  i 

-6 

(b) 

-8 

-10 

-12 

1 

-14 

(a)< 

X-6 

y  =  l 

5 

-1fi 

/ 

-18 

f 

-20 

-22 

-24 

^=* 

CHAPTER  XVI 
NON-LINEAR  ALGEBRAIC  EQUATIONS 

1.  The  first  degree  equation  (1)  is  called  a  linear  formula 
from  the  fact  that  it  corresponds  to  a  straight  line  graph 
when  plotted  to  two  rectangular  axes. 

(1)  y  =  a+bx. 

(2)  yi  =  bx. 

We  have  observed  that  (2)  is  a  special  form  of  (1)  when 
the  a  term  becomes  zero.  (1)  and  (2)  have  the  same  slope 
and  are  therefore  parallel  lines.  If  the  distance  a  is  added 
to  (2)  then  (1)  may  be  written  (3), 

(3)  y  =  &+yi. 

The  interpretation  of  (3)  states  that  the  ordinates  of 
curve  (1)  are  a  greater  than  the  ordinates  of  (2).  The 
two  elements  which  change  the  position  of  a  line  are  its 
slope  b  and  its  vertical  axal  intercepts  a. 

2.  A  non-linear  equation  is  one  which  differs  from  (1) 
in  some  degree,  i.e.,  one  whose  graph  is  not  a  straight  line 
as  illustrated  in  the  algebraic  and  non-algebraic  equations 

(4)  .  .  .  (17). 

316 


NON-LINEAR  ALGEBRAIC  EQUATIONS  317 


ALGEBRAIC. 

NON-ALGEBRAIC. 

(4) 

y  =  a+bxK 

(11)  2/  =  sin  x. 

(5) 

y  =  a-\-bx2. 

(12)  y  =  log  x. 

(6) 

y  =  bx3. 

(13)  y  =  a-\-b  sin  a:. 

(7) 

y  =  a-\-bxn. 

(14)  y  =  sm~1(x). 

(8) 

y  =  a+bxn+cxm. 

(15)  y  =  e. 

(9) 

y  =  bx±m. 

(16)  y=eax  sin  (bx+c). 

(10) 

y  =  a-\-bx-{-cx2-\-dx3-\-cx4-\-.  . 

.     (17)  #  =  82. 

Algebraic  equations  involve  only  the  operations  of 
addition,  subtraction,  multiplication,  division,  and  com- 
mensurable powers  of  the  variable. 

3.  Parabolic  Curves.  The  equation  y  =  xm  represents 
a  standard  parabola.  There  are  infinite  number  of  para- 
bolic curves  obeying  the  law  that  the  ordinate  ;s  equal  to 
a  definite  power  of  the  abscissa.  For  every  positive  integral 
or  positive  fractional  power  of  x  there  is  a  definite  standard 
parabola.  Plot  the  graphs  for  y=xm  for  the  following 
values  of  the  exponent  m. 

Curve  of  Cube  Roots.  Curve  of  Solenoid  Winding. 

_    ' '  12  odd  __       _  4  even 

Ex.1.    »«-«r-rr(  Ex.    6.    w=--r-r. 

3     6  odd  3  odd 

Curve  of  Square  Roots.  Curve  of  Carrying  Capacity. 

„     n            1     2  odd  _       _  3     1Eodd 

Ex.2.    m=-=- .  Ex.    7.    m=-=1.5 


2  4  even  2  even 

Curve  of  Fusing  Effects  of  Currents.  Curve  of  Squares  of  Copper  Losses. 

_  •  > -v  2     4  even  __     _  _     2  even 

Ex.3.    m=-=--rr.  Ex.    8.    771=2=--^. 

3  6  odd  1  odd 

Curve  of  Cubes  or  Luminous  Intensity. 

3       ^    odd  _        _  _     3  odd 

Ex.4.    m=-=.75 .  Ex.    9.    w=3  = 


4  even  1  odd 

Curve  of  Energy  Radiation  of  Black  Body. 

„  1   odd  _      . .  .  even 

Ex.  5.     m  =  1  =  -  -vr-  Ex-  10-     m  =4  -JT- 

.1   odd  odd 


318  PRACTICAL  MATHEMATICS 

Prepare  and  label  a  table  of  values  for  x  and  y  in  each  example. 
Assume  ten  values  for  x  and  determine  the  corresponding  values 
for  y.  The  values  must  be  so  chosen  that  they  may  be  located 
on  the  axes  of  the  cross-section  paper.  The  origin  of  the  axes 
should  be  located  at  the  center  of  the  paper  and  equal  scales  chosen 
for  both  axes.  Plot  the  ten  curves  on  the  same  sheet  of  paper 
to  the  same  scale.  Arrange  a  blank  sheet  of  paper  to  contain  the 
ten  tables  and  place  this  sheet  so  as  to  face  the  graphs.  Letter  or 
number  each  graph  to  correspond  to  a  like  letter  or  number  which 
appears  in  the  label  of  the  table  so  as  to  facilitate  cross-reference. 

4.  Fig.  121  represents  the  graphs  of  y  =  xm  for  the  follow- 
ing values  of  m:  J,  ^,  J,  1,  2,  3,  4,  which  correspond  to 
graphs  (/),  0),  (d),  (g),  (a),  (b),  and  (c)  respectively. 

Each  parabola  is  divided  at  the  origin  into  two  equal 
parts.  Curve  .(c),  which  has  the  greatest  exponent  shows 
the  least  divergence  from  the  Y  axis,  whereas  curve  (/), 
which  has  the  least  exponent,  shows  the  greatest  divergence 
from  the  Y  axis.  The  numeric  order  of  the  exponents  is 
also  the  numeric  order  of  the  graphs,  (g)  is  recognized  as 
a  straight  line  and  separates  the  graphs  which  have  integral 
exponents  from  the  graphs  which  have  decimal  exponents. 

Every  parabola  passes  through  the  common  point  (1,1) 
which  is  obtained  for  each  equation  by  substituting  x  =  1  in 
y  =  xl.  Why?  The  area  near  the  origin  is  enlarged,  as 
shown  in  the  upper  group  of  graphs  in  Fig.  121,  which  may 
be  readily  identified  by  their  letters.  In  each  case  half  of 
the  parabola  is  located  in  the  first  quadrant.  The  second 
half  of  the  parabola  may  be  located  in  the  second,  third,  or 
fourth  quadrant,  depending  upon  whether  the  exponent 
reduces  to  a  ratio,  between  an  even  and  an  odd  number, 
between  two  odd  numbers,  or  between  an  odd  and  an  even 
number.  A  quantity  with  a  fractional  exponent  may  be 
written  in  the  radical  form.  The  denominator  of  the 
exponent  becomes  the  index  of  the  root  and  the  numerator 
of  the  exponent  becomes  the  power  index  of  the  quantity. 
Thus 

(18)  y=x5=^x3    and    y  =  xi  =Vx\ 


NON-LINEAR  ALGEBRAIC  EQUATIONS  319 


— c 

) 

:-o!s 

3 

i  f 

4  1 

& 

Is 

^(1,1)       Enlargement  of  the  Area 

near  the  origin,  which  shows 
y>"  that  all  curves  of  the  type  Y=X' 
/       Bass  through  the  common  point 

>    i| 

"^i 

X  x  x  x  x  x  x  x 

Il    II  III  *  II   II  II  II 

S  S3  3  SS>3 

s 

m 

ESj       A 
St 

3J 

s_ 

S^, 

ol      1 

$ 

J  1 

LA 

*               : 

0 

*i 

\\ 

il 

s 

^ 

-16 

*»  o — 

3 

cP=< 

..,___ 

3                             - 
3                             . 

1                    : 

5              a 

f        J 

a 1 

s          i 

>            s 

5  §. « 

s                 ; 

5           n   * 

1              3 

1 

3^ 

' 

-o-f 

>■ 
1 

-£>. 

\$ 

ft 

"1 

\ 

> 

1 
C  o 

I 

1 

320 


PRACTICAL  MATHEMATICS 


Table  XXII  shows  the  quadrant  positions  of  the  two 
halves  of  a  parabola  for  the  three  types  of  exponents. 

TABLE  XXII.     QUADRANT  POSITIONS  OF  y  =  x±m 


Reduced 
Exponent 

Equivalent 

Root  and 

Power. 

Assigned  Values  of 

Quadrant 

Positions  of 

Graphs. 

Axis 
of  Sym- 

of X. 

X 

V 

metry. 

even 
oddf 

+ 

+ 
+ 

II         I 

even     odd 

odd^even 

Y 

odd 
odd 

+ 

+ 

I         III 

odd      odd 

^^odd 

odd 
even 

+ 
+ 

+ 

I        IV 

odd      even 

^V*odd 

X 

5.  Families  of  Curves.  A  family  of  curves  is  a  group 
of  graphs  which  is  plotted  by  varying  a  parameter,  i.e., 
one  of  the  arbitrary  constants  in  an  equation.  The  varia- 
tion of  m  in  y  =  bxm  gives  a  family  of  parabolas  which  are 
represented  partially  in  Fig.  121 .  The  variation  of  b  gives  a 
family  of  parabolas  which  are  represented  partially  in 
Fig.  122. 

Ex.  11.  Plot  y=bx2.  Prepare  and  label  five  tables  for  the 
corresponding  graphs,  when  fe  =  .l,  .5,  1,  2  and  10.  Describe  the 
effect  of  changing  the  coefficient  b. 

6.  Fig.  122  shows  a  family  of  odd-odd  parabolas  obtained 
by  varying  the  coefficient  of  b  in  y  =  bx3'  .  II  is  the  standard 
parabola  for  which  6  =  1.  Curves  I,  III,  IV,  V  show  the 
effect  of  changing  b  and  correspond  to  the  graphs  of 
y  =  bx*8,  when  b  equals  2,  .5,  .25,  and.l  respectively. 

7.  Fig.  123  shows  a  family  of  even-odd  parabolas 
obtained  by  varying  b  in  the  equation  y  =  bx2A.  The 
standard     parabola    is    indicated    as    ©-(£).      The    other 


NON-LINEAR  ALGEBRAIC  EQUATIONS  321 


Fig.  122.— Odd-odd  Parabolas  y  =bx 


322 


PRACTICAL  MATHEMATICS 


®6=.l  ®&=2 

©b  =.2  ©6=1 

®&=.3  ©6=-5 

©6=4  ©6=.4 

§6=.5  @6=.3 

b=  1  ©  &=.2 

6=2  U4)6=.l 


Fig.  123. — Even-odd  Parabolas. 


NON-LINEAK  ALGEBRAIC  EQUATIONS  323 

parabolas  are  identified  by  the  number  index  which  gives 
the  value  of  the  parameter  b.  The  exponent  2.4  =  fo  =  V2, 
which  comes  under  the  even-odd  class.  Curves  of  this 
class  are  symmetrical  to  the  Y  axis,  i.e.,  on  every  perpen- 
dicular to  the.  Y  axis  there  is  a  pair  of  alternate  points 
equally  distant  from  the  axis.  The  axis  cuts  a  parabola 
at  its  vertex. 

Ex.  12.     Plot  the  data  from  Ex.  18,  Chapter  IX,  which  corre- 
sponds to  equation, 

P/^.003551-6. 

Ex.  13.    Plot  the  following  equations  on  one  sheet  of  paper  to 
the  same  scale  and  axes. 


(19) 

2/i  =  ex2, 

(20) 

y2=a+cx2, 

and 

(21) 

y3=a+bx+cx2, 

8.  The  Composition  of  Curves  by  Addition  and  Sub- 
traction. The  interpretation  of  (20)  states  that  the  ordi- 
nates  of  2/2  exceed  the  ordinates  of  2/1  by  the  constant 
amount  a.  The  graph  of  (20)  will  be  a  duplication  of  (19), 
but  the  curve  will  be  moved  vertically  upward  a  distance 
a  without  alteration  in  shape.  Substituting  (20)  and  (22) 
in  (21)  we  obtain  (23). 

(22)  1/4  =  bx. 

(21)      y3  =  a-\-bx+cx2.  /.     (23)     2/3  =  2/2+^  =  2/2 +2/4. 

(22)  is  the  equation  of  a  straight  line.  The  interpre- 
tation of  (23)  states  that  the  graph  for  2/3  may  be  obtained 
by  adding  the  corresponding  ordinates  of  graphs  (20)  and 
(22) .  Check  the  result  from  calculation  by  adding  the  ordi- 
nates graphically  with  a  bow  dividers.  What  change  has 
taken  place  in  the  position  of  the  axis  of  the  graph  (21) 
compared  with  (19)  and  (20). 


324  PRACTICAL  MATHEMATICS 

9.  Hyperbolic  Curves.  The  equation  y  =  x~m  repre- 
sents a  standard  hyperbola.  There  are  an  infinite  number 
of  hyperbolic  curves  obeying  the  law  that  the  ordinate  is 
equal  to  the  reciprocal  of  a  definite  power  of  the  abscissa. 
For  every  negative  integral  or  fractional  power  of  x  there 
is  a  definite  standard  hyperbola. 

<24)  y=x~m=^ 

The  interpretation  of  (24)  states  that  for  every  definite 
assigned  value  of  m  the  value  of  the  ordinate  of  the  standard 
hyperbola  is  numerically  the  reciprocal  of  the  ordinate 
of  the  corresponding  standard  parabola.  The  tables  for 
plotting  hyperbolas  may  be  very  readily  obtained  from  the 
tables  for  plotting  parabolas.  In  both  tables  the  decimal 
and  integral  values  of  x  are  the  same  but  the  values  of  y  in 
the  table  for  hyperbolas  are  obtained  by  reciprocating  the 
values  of  y  in  the  corresponding  table  for  parabolas.  All 
standard  hyperbolas  pass  through  the  common  point  (1,  1). 
The  hyperbolas  have  their  two  halves  called  branches 
distinctly  separated.  The  quadrant  positions  of  hyper- 
bolas are  obtained  from  Table  XXII. 

10.  Fig.  124  shows  three  odd-odd  parabolas  obtained  by 
plotting  y  =  bx1A.  Graphs  (a),  (b),  and  (c)  correspond  to 
the  respective  values  1,  2,  .5  for  the  parameter  b. 

11.  Fig.  125  shows  three  odd-odd  hyperbolas  obtained 
by  plotting  y  =  bx~1A.  The  numeric  values  of  the  ordinates 
of  (a),  (6),  (c),  in  Fig.  125,  are  the  reciprocals  of  the  numeric 
values  of  the  ordinates  of  (a),  (6),  (c),  respectively  in 
Fig.  124.  A  point  indefinitely  near  the  origin  in  Fig.  124 
is  indefinitely  removed  from  the  origin  in  Fig.  125. 

Integral  values  in  the  one  family  of  curves  are  decimals 
in  the  other  family  of  curves. 

12.  Powers,  Roots,  and  Reciprocals.  Parabolas  may  be 
used  to  obtain  powers  and  roots  of  numbers  and  hyperbolas 


NON-LINEAR  ALGEBRAIC  EQUATIONS  325 


1 

9- 

Y=bxu 

Y 

J 

(a)b=   1 
(6)6=  2 

(c)6  = 

M 

6 

5 

(&)/ 

<«)/ 

\*y 

2 

rf 



-X 

1 

X 

i 

1 

y-f- — 1 

( 2 

1 

1 

1 

1 

/c) 

3 

/(*) 

M 

5 

6 

-Y 

9- 

Fig.  124.— Odd-odd  Parabolas. 


326 

PRACTICAL  MATHEMATICS 

q 

Y8 

< 

V  =  bxhi 
(a)  6=1 
(6)  6  =  2 
(c)  6=y2 

1 

lb 

in 

\ia) 

-X 

: 

) 

! 

^te 

X 

^s 

!         ; 

3 

. 

)           t 

(&K 

C\C) 

2 

i 

3 

1 ' ' 

1 1 

-Y 

r 

< 

9 

Fig.  125.— Odd-odd  Hyperbolas. 


NON-LINEAR  ALGEBRAIC  EQUATIONS  327 

may  be  used  to  obtain  reciprocals  of  numbers,  reciprocals  of 
powers  and  reciprocals  of  roots  of  numbers. 

13.  Fig.  126  shows  a  family  of  odd-even  hyperbolas 
plotted  from  y  =  bx~13  in  whch  the  values  of  b  are  .1, 
.5,  1,  2,  10.  The  curves  are  identified  by  the  reference  index 
of  lines  showing  light,  heavy,  solid,  dash,  and  dot-dash 
lines.  Compare  the  family  of  parabolas  of  Fig.  124  with 
the  family  of  hyperbolas  of  Fig.  125  by  noting  their  rela- 
tive position  toward  the  Y  axis.  What  is  the  change  in 
position  of  hyperbolic  graphs  when  the  parameter  b  is 
increased  or  decreased. 

14.  Other  non-linear  algebraic  equations  are  repre- 
sented in  (26),  (27),  (28),  and  (29),  which  are  known  as 
power  series. 

(25)  y  =  a+bx+ca?+da?+e&+.  .  .  +kxm. 

All  composite  parabolic  equations  are  modified  forms 
of  (25)  and  (26)  from  which  they  may  be  obtained  by 
assigning  definite  values  to  the  coefficients  a,  b,  c,  h,  e,  etc. 

(26)  yn  =  a+bx+cx2+dx2+ex4  +  .  .  .+kxm. 

The  degree  of  x  in  (25)  and  (26)  is  numerically  the 
same  as  ,the  highest  exponent  m  in  any  of  its  terms.  (25) 
is  a  single  valued  function  of  y  whereas  (26)  is  a  multivalued 
function  of  y.  For  every  single  value  of  y  in  (25)  there  will 
be  m  values  of  x.  Such  an  equation  when  plotted  will 
represent  a  graph  which  rises  and  falls  with  no  regularity. 
The  number  of  crests  and  troughs  in  the  graph  will  be 
one  less  than  the  degree  of  the  equation.  At  the  crest  of 
any  graph  there  will  be  a  maximum  point  whose  ordinate 
is  numerically  greater  than  the  ordinates  of  the  contiguous 
points.  At  the  trough  of  any  graph  there  will  be  a  minimum 
point  whose  ordinate  is  numerically  less  than  the  ordinates 
of  the  contiguous  points. 


328 


PRACTICAL  MATHEMATICS 


Y8 

1 

w 

111 

,6=10 

,6=2 

,6=1 

,6  =.5 

,6=.l 

.3 

! 

W 
i 

1 1 

nil 

3 

1 

mi 

I 

te 

^ 

1        X 

Xfl 

i 2rrrr=3===== 

fe=d 

^ir^y- 

fi=^===3 

_=.-■= 

; 

r"    ^^jj 

i 

1       /     c 

-3 

1 

1   ,' 
III 

1 

-5 

1 

f 

• 
1 

-Yn 

t 

-!) 

i 

"Wra    19fi— OrM— p.vpn  HvnprhnlnK. 


NON-LINEAR  ALGEBRAIC  EQUATIONS  329 

In  Fig.  170  P  and  U  are  maximum  points  and  A  is  a 
minimum  point.  The  interpretation  of  (26)  states  that 
there  are  n  different  values  of  y  which  will  produce  the 
same  set  of  values  of  x.  As  an  illustration  let  n  =  2,  a  =  6.25, 
c=—  .25,  and  all  other  coefficients  equal  zero.  After  sub- 
stitution and  transposition  (26)  reduces  to  (1)  of  Ex.  17, 
which  is  the  ellipse  shown  in  Fig.  129.  For  every  value 
of  y  (+  or  — )  there  are  two  values  of  x  and  likewise 
for  every  value  of  x  ( +  or  — )  there  are  two  values 
of  y.  Parabolas  with  integral  exponents  are  forms  of 
(25),  whereas  parabolas  with  fractional  exponents  are 
forms  of  (26). 

Other  power  series  are  given  by  (27)  and  (28)  of  which 
the  hyperbolas  are  corresponding  forms, 

/o*7\  i  6  i   c  i  d  ,   e  . 

(27)  y=a+x+x^+^+  ■'- 

(28)  ^v«+|+|+|+£+  .... 

In  order  to  abbreviate  the  right  hand-members  of 
(25)  .  .  .  (28),  we  use  the  symbol  f(x),  which  reads 
f  function  of  x.  A  function  is  a  quantity  which  depends 
upon  another  quantity  for  its  value.     Since  y  in  (25)  .  .  . 

(28)  depends  upon  the  value  of  x  then  the  left-hand  members 
of  the  equations  equal  functions  of  x.  Other  distinguishing 
symbols  follow. 

Therefore  (25)  .  .  .  (28)  may  be  abbreviated  respect- 
ively as  (29)  .  .  .  (32). 

(29)  y=fi(x).  (30)  y=fs(x)     or    y  =  4>(s). 

(31)  y=f2(x)  or  y  =  F(x).  (32)  y=U(x)     or    y  =  ^(x). 

The  symbol  for  function  may  be  indicated  by  /,  or  /  with 
a  subscript,  or  by  <j>  or  6,  etc.     The  dependent  variable  is 


330  PRACTICAL  MATHEMATICS 

i 
always    a   function    of   the   independent   variable.     What 
functions  of  x  are  represented  in  y  =  sin  x  and  y  =  log  x. 

15.  Conic  Sections.  A  quadratic  equation  in  y  and  x 
may  contain  the  first  and  second  powers  of  both  y  and 
x,  a  product  term  xy,  and  a  constant  or  absolute  term. 
Such  a  complete  quadratic  equation,  i.e.,  one  which  con- 
tains all  the  mentioned  terms  is  represented  in  (33), 

(33)  ax2+bxy+cy2+dx+ey+f=0. 

There  are  four  types,  i.e.,  four  distinct  families  of  curves 
corresponding  to  (33).  They  include  circles,  ellipses,  conic 
parabolas,  and  conic  hyperbolas,  and  are  known  as  conic 
sections.  They  were  recognized  by  the  Greeks  as*  the  sec- 
tions of  cylinders,  spheres,  and  cones. 

16.  Fig.  15,  page  61,  shows  how  the  conic  sections  may 
be  obtained  from  a  right  circular  cone,  i.e.,  a  cone  having 
a  circular  base  with  its  apex  located  vertically  above  the 
center  of  the  base.  When  a  cutting  plane  is  parallel  to  the 
base  the  section  will  be  a  circle,  such  as  MNPQ.  If  the 
cutting  plane  is  rotated  slightly  so  as  to  be  non-parallel  to 
the  base,  the  section  becomes  an  ellipse,  such  as  DBEC. 
If  the  angle  of  the  cutting  plane  be  further  increased  the 
ratio  of  the  major  and  minor  diameters  of  the  ellipse  is 
increased.  When  the  cutting  plane  is  rotated  further  so 
as  to  be  parallel  to  the  element  0  10  the  section  becomes 
a  parabola,  such  as  KFCGL.  If  the  angle  of  the  cutting 
plane  be  further  increased  the  cutting  plane  will  cut  both 
nappes  of  the  cone  and  the  resulting  sections  become  hyper- 
bolas. 

17.  Conic  Test.  A  study  of  the  effect  of  changing  the 
values  of  the  coefficients  of  (33)  leads  to  a  precise  way 
of  determining  the  conditions  which  result  in  a  curve  of 
each  type.  The  numeric  values  of  a,  b,  and  c  are  sufficient 
to  determine  the  kind  of  conic  according  to  the  following 
relations.    All  terms  of  a  quadratic  equation  must  be  trans- 


NON-LINEAR  ALGEBRAIC  EQUATIONS  331 

posed  to  one  member  of  the  equation,  as  shown  in  (33), 
and  then  A  is  the  coefficient  of  the  x2  term,  C  is  the  coeffi- 
cient of  the  y2  term,  and  B  is  the  coefficient  of  the  xy  term. 
Multiply  4A  by  C  and  compare  it  with  Bl  as  follows : 

When  B2<4AC  the  equation  (33)  represents  an  ellipse. 

When  B2  =  4AC  the  equation  (33)  represents  a  parabola. 

When  B2>4AC  the  equation  (33)   represents  an  hyper- 
bola. 

When  B2  =  0  and  A  =  C  the    equation  (33)  represents  a 
circle. 

When  B  =0  then  B2  is  greater  than  any  negative  value  of 
the  product  4AC. 

Ex.  14.  Apply  the  conic  test  and  substantiate  the  designation 
of  the  following  simplified  forms  of  conic  formulas,  r  is  the 
radius  of  the  circle,  a  and  b  are  the  respective*  semimajor  and 
semiminor  axes.  H  and  K  are  the  respective  coordinates  of  the 
center  of  an  eccentric  conic,  i.e.,  one  whose  center  is  not  at  the 
origin,     c,  d,  k,  m,  n,  are  arbitrary  constants. 

(34)  x2-\-y2  =r2.         The  equation  of  a  circle 
whose  center  is  at  the  origin. 

(34A)  x2  +y2  =  25.      What  is  the  value  of  r? 

(34B)     (x  —H)2+(y  —K)2  =r2.     The  equation  of  an  eccentric  circle. 
See  Figs.  127  and  130. 

(35)  (x  -5)2  +  (y  -6)  =49.    What  are  the  values  of  r,  H,  and  K. 

(36)  (x  -3) 2  +  (y  +3)  =  16.      What  are  the  values  of  r,  H,  and  K. 

There  will  be  one  definite  circle  corresponding  to  a  definite 
value  assigned  to  the  radius  r. 

A  circle  may  be  moved  about  in  a  plane  by  changing  the 
numeric  values  of  the  coordinates  of  its  center.     The  equation  (36) 


332 


PRACTICAL  MATHEMATICS 


holds  true  for  any  position  in  the  four  quadrants,  as  may  be  shown 
from  the  relations  of  the  sides  of  the  right  angle  triangles. 


Fig.  127.— The  Circle  (x  -  H)2  +  (y  -  K)*  =r\ 


<37>    a'+T^ 


The  equation  of  an  ellipse. 
See  Figs.  129  and  133.    The  center  is  at  the  origin. 


(x—H)2     (v—K)2 

(38)      — +  — =  1.     The  equation  of  an  eccentric  ellipse. 

v     '  a2  b2 


NON-LINEAR  ALGEBRAIC  EQUATIONS  333 

For  every  definite  pair  of  values  of  a  and  b  there  is  a  definite 
ellipse.  An  ellipse  may  be  moved  vertically  and  horizontally  in 
the  plane  by  changing  the  respective  coordinates  of  its  center.  A 
family  of  ellipses  which  have  common  foci  are  confocal. 

(39)  ^  +^  =  1 .      What  are  the  values  of  a  and  6? 

t:»7        dO 


(40)  —  +—  =  1 .      What  are  the  values  of  a  and  6? 
oo     49 

What  comparison  is  there  between  the  ellipses  (39)  and  (40)? 

(41)  (*~3)2  +  ^~4)2  =  l-    What  are  the  values  of  a,  b,  H,  and  K? 

do  25 


(42)  —  -—  =1.  The  equation  of  an  hyperbola. 

See  Fig.  135.     The  center  is  at  the  origin. 

(x—H)2     (y—K)2 

(43)    - — — —  =  1 .  The  equation  of  an  eccentric  hyperbola. 

For  every  definite  pair  of  values  of  a  and  b  there  is  a  definite 
hyperbola.  An  hyperbola  may  be  moved  vertically  and  hori- 
zontally in  the  plane  by  changing  the  respective  coordinates  of  its 
center.    Confocal  hyperbolas  have  common  foci. 

x2     v2 

(44)  —  —  —  =  1.  What  are  the  values  of  a  and  b? 
49    do 

x2     y2 

(45)  —  —  —  =  1,  What  are  the  values  of  a  and  6? 
do    49 

(46)  (£^2-(y27<f'5)2==L  What  are  the  values  of  a,  6,  F,  and  X? 

(47)  (*  -2) »  -  (y +1)»  - 1 .       What  are  the  values  of  a,  b,  H,  and  K? 

(48)  y  =kx2.  The  equation  of  a  parabola. 


334  PRACTICAL  MATHEMATICS 

The  vertex  is  at  the  origin  and  the  curve  is  symmetrical  to  the 

k 
Y  axis.    Its  focus  is  —  units  from  the  vertex. 
4 


(49)   x  =my2.  The  equation  of  a  parabola. 

The  vertex  is  at  the  origin  and  the  curve  is  symmetrical  to  the 

TYt 

X  axis.    Its  focus  is  —  units  from  the  vertex. 
4 


Describe  the  parabo] 

[as 

(50)            y=8x*, 

(51) 

x  =  %», 

(52)             y=-4x2 

and 

(53) 

x=-16y\ 

The  diameter  of  a  parabola  is  that  portion  of  the  axis  lying 
between  the  vertex  and  last  ordinate  of  the  curve.  The  entire 
parabola  and  its  entire  diameter  cannot  be  represented  as  they 
extend  indefinitely  from  the  vertex. 

(54)  y  =  {c+kx)\ 

The  equation  of  a  parabola  whose  vertex  is  located  c  units 
vertically  above  the  origin  but  otherwise  like  (48). 

(55)  x  =  (d+my)*. 

The  equation  of  a  parabola  whose  vertex  is  located  d  units 
horizontally  to  the  right  of  the  origin  but  otherwise  like  (49). 
Describe  the  parabolas  (56A)  y  =  (1  +Sx) 2,  (56B)  y  =  ( - 1  -  8x) 2, 

(56C)         z  =  (l+4?/)2,  (56D)  x  =  (-2+8y)\ 

Write  equations  for  a  circle,  ellipse,  parabola,  and  hyperbola, 
using  the  data:  r=4,  a=S,  6=4,  H  =  1,  K=2,  k=6,  c  =  1.5, 
m  =  10,  d  =2.5.    Describe  each  curve. 

Plot 

(57)  x*+y*=a\ 


NON-LINEAR  ALGEBRAIC  EQUATIONS 


333 


a  to  have  a  special  value  according  to  the  following  assignment: 
0-1,  1.5,2,2.5,3,3.5,4,4.5,5. 
Plot 


(58) 


X-+y-=l. 


a  and  b  are  to  have  specialty  assigned  values  in  pairs  as  follows: 


a 

i 

i 

2 

2 

3 

4 

3 

3 

2 

1 

b 

2 

3 

3 

4 

4 

3 

2 

1 

1 

1 

Plot 


(59) 


a2      fc2 


a  and  6  are  to  have  specially  assigned  values  the  same  as  above. 
Plot 


(60) 


y=kx2  and  (61) 


x=myi, 


upon  the  same  sheet  of  paper.    Use  the  same  pair  of  axes  for  both 
curves  and  assume 

&=4=m. 

What  significant  relation  do  these  equations  bear  to  each 
other?  What  mechanical  operation  is  equivalent  to  interchanging 
x  and  y  in  the  equation? 


The  meaning  of  the  xy  term,  equations  (34)  to  (61), 
represent  standard  curves  which  contain  no  xy  term  and 
for  which  corresponding  graphs  have  their  diameters  parallel 
to  the  axes.  The  introduction  of  the  xy  term  has  the  effect 
of  rotating  the  diameter  of  a  conic  through  an  angle  0° 
from  its  normal  position.    The  angle  of  rotation  is  measured 


336  PRACTICAL  MATHEMATICS 

by  the  tangent  as  expressed  in  (62)  in  which  A,  B,  and  C 
are  the  coefficients  of  (33) . 

(62)  tan  20 = j— g. 

Ex.  15.    Describe  the  curve  and  its  position  and  then  plot  the 
following  equations: 

clx 

(63)  y= r>  when  a=2.5,  c=3,  d  =  .5. 

c+dx 

(64)  y  =  {a+bx)~x,        when  a  =  1,6=8. 

(65)  xy  =  ax  +by,  when  a  =  \,  6=5. 

(66)  ch  =  T\  when  T  =  1.5. 

(67)  A  =p(l  +j~\ ,      when  A  =  100. 

(68)  —  =a+bp,  when  a  =  l,  6  =  .5. 

(69)  #=a-l — ,  when  a  =  1,  6=5. 

(70)  h  =^,  when  &«.2,  / =25,  v  =  10,  d  =75. 

(71)  (y-x)(x+2y-3)=7. 

(72)  5z2+2a:?/+5?/2-12z-12?/=0. 

(73)  x2  -2xy  +y2-8x +16=0. 

The  solution  of  non-linear  simultaneous  equations  is  the 
determination  of  the  pairs  of  values  of  the  variables  which 
will  satisfy  the  given  equations.  The  solution  may  be 
determined  graphically  by  observing  the  coordinates  of  the 
points  of  intersection  of  the  graphs. 


NON-LINEAR  ALGEBRAIC  EQUATIONS  337 

In  general,  two  or  more  simultaneous  equations  have 
as  many  solutions  as  the  product  obtained  by  multiplying 
the  degree  of  each  equation.  This  is  substantiated  in  the 
graphic  work  which  follows. 


U2  -8  -4  V  ■«  8  12  ly  20 

x    Fig.  128. — Graphic  Solution  of  Simultaneous  Equations. 

Solve  graphically  the  following  simultaneous  pairs  of 
equations : 

Ex.  16.     (a)        x-Vy=0. 

(b)        x+y=20.    See  Fig.  128. 

The  intersections  give  (-4.8,  24.8)  and  (4,  16). 


338 


PRACTICAL  MATHEMATICS 


Equation  (a)  represents  a  parabola  and  being  devoid  of  a 
constant  term  the  vertex  is  at  the  origin.  Equation  (b)  repre- 
sents a  line  whose  slope  is  (-1)  and  whose  X  intercept  is  (+20). 

Ex.17.     (1)  x2+4?/2=25. 

(2)  2xy=12. 


t> 1 

[l 

(1)  X2+4Y2=  25  ELLIPSE 

(2)  2XY-1  2  HYPERBOLA 

S        t 

A  2) 

2 

C 

0s 

^4,1. 

)) 

F. 

H 

M 

7       - 

6      - 

5 

4       - 

3       - 

2       - 

1 

1 

2 

3 

4       j 

'5 

6 

A 

&£~4,~ 

1.5) 
£8.- 

2) 

1 

B 

2 

•fl) 

Q        £ 

\») 

3 

4 

K 

5 

Fig.  129. — Non-linear  Simultaneous  Equations. 

The  four  solutions  are  (3,  2),  (4, 1.5),  (-3,  -2),  and  (-4,  -1.5). 
Why  are  there  four  solutions?  Are  the  graphs  normal?  See 
Fig.  129. 


Ex.  18.     (a) 

n 


4(x+y)=3xy. 
z+2/+z2+?/2=26. 


NON-LINEAR  ALGEBRAIC  EQUATIONS  339 

These  equations  are  plotted  in  Fig.  130.    Describe  the  graphs 
and  state  the  solutions  of  the  equations. 


Ex.  19.     (1) 

(2) 


x3+y3  =65. 


Fig.  130. — Two  Simultaneous  Quadratics  Showing  Four  Solutions. 

How   many   solutions   should   there    be   for   these   equations? 
State  their  values.    See  Fig.  131. 

Ex.  20.     Plot 
(71)  4x*+4xy+y2-8x-4:ly+±00=0 


340 


PRACTICAL  MATHEMATICS 


on  the  same  cross-section  paper  with  the  graphs  of  Ex.  16.  It  will 
be  observed  that  the  above  equation  may  be  reduced  to  (75)  and 
(76). 


(75) 
(76) 


(2x+y-20)*=y. 

2x+y-20  =  Vy. 


8 
7 

6 

5 

(Dx 

(2)> 

4-  Y  =  5 

•3+  y3=  e 

STRAIG 
5  CUBI 

HT  LIN 

E 

A 

4    \ 

jCU) 

D        B^ 
3 

2 

1 

(4,1) 

c' 

r*    \ 

E 

i 

3 

2 

1 

1 

1 

2 

3      i 

0 

(2) 

7 

2 

3 

\ 

4 

\ 

9 

Fig.  131. — Line,  Linear  and  Non-linear  Simultaneous  Equation. 

Equation  (76)  is  the  sum  of  (a)  and  (6)  in  Ex.  16. 
Ex.  21.     Plot 
(77)  x2-3xy+y*+5x+5y-2Q=0, 


NON-LINEAR  ALGEBRAIC  EQUATIONS  341 

on  the  same  cross-section  paper  with  the  graphs  of  Ex.  18.    Add 
the  equations  of  Ex.  18  and  compare  with  Ex.  21. 
Ex.  22.     Plot 

(78)  x>-xy+y*  =  13, 

on  the  same  cross-section  paper  with  the  graphs  of  Ex.  19.     Divide 
(2)  by  (1)  in  Ex.  19  and  compare  with  Ex.  22. 


CHAPTER  XVII 


ECCENTRICITY  OF  CONICS 


1.  The  ellipse,  parabola,   and  hyperbola  may  be  con- 
structed with   the  aid  of   the   straightedge    and    compass 
without  the  necessity  of  plotting 
points  from  calculations. 

The  conies  mentioned  above 
have  the  common  property  that 
every  point  on  each  curve  pre- 
serves a  constant  ratio  between 
its  distance  from  a  fixed  point 
(focus  or  pole)  to  the  distance 
from  a  fixed  line  (directrix). 
These  distances  are  designated 
as  the  focal  and  directral  dis- 
tances respectively.  In  Fig.  132, 
P  is  a  point  on  the  curve.  F  is  the  focus, 
directrix. 


Fig.  132—  The  Analytic 
Definition  of  a  Conic. 


DD'  is  the 


(1) 


Focal  distance 


PF 
PH 


Directral  distance 

=  Vy2  +  (x-m)2 
x 


=  e  =  a  numeric  constant. 


2.  The  fixed  ratio  e  is  called  the  eccentricity  of  a  conic. 
For  any  one  definite  curve  e  is  fixed  in  value,  i.e.,  a  numeric 
constant,  but  e  may  have  any  positive  value.  For  all  values 
less  than  one  (e  <  1),  e  determines  an  ellipse,  and  for  all  values 
of  e  in  excess  of  one  (e>  1),  e  determines  an  hyperbola.  When 
e  =  1,  then  the  conic  is  a  parabola. 

342 


ECCENTRICITY  OF  CONICS  343 

The  parabola  lies  at  the  boundary  between  ellipses  and 
hyperbolas. 

The  conic  test  observed  in  the  last  chapter  is  consistent 
with  the  following  order  of  magnitudes: 

Ellipses  are  constructed  for  values  of  e  <  1 ; 
Parabolas  are  constructed  for  values  of  e  =  1 ; 
Hyperbolas  are  constructed  for  values  of  e>  1; 
Circles  are  not  constructed  for  values  of  6  =  0; 
since  the  directral  distance  is  infinite. 

The  directral  distance  PH  is  measured  perpendicular  to 
the  directrix  DD' . 

The  focal  distance  is  measured  radially  from  the  focus  F. 
The  focus  is  located  at  a  distance  m  from  the  directrix, 
and  the  vertex  lies  between  the  directrix  and  the  focus.  When 
the  eccentricity  of  a  conic  is  known  the  equation  for  the 
graph  may  be  obtained  by  substituting  the  values  of  e  and 
m  in  (1)  or  in  its  simplified  form  given  in  (2). 

(2)  (l-e2)x2-2mx+y2+m2  =  0. 

3.  The  ellipse,  Fig.  133,  the  parabola,  Fig.  134,  and  the 
hyperbola,  Fig.  135,  may  be  constructed  with  the  aid  of  the 
straightedge  and  compass  without  the  necessity  of  plotting 
points  from  calculations. 

In  each  case  the  directrix  is  shown  as  a  vertical  line. 
The  horizontal  axis  is  drawn  through  a  point  0  on  the 
directrix  and  through  the  focus  F.  The  diagonal  lines 
OS  and  OS'  are  drawn  so  that  their  slope  equals  e,  which 
is  the  numeric  value  of  the  eccentricity  of  the  desired  conic. 
Then  any  perpendicular  drawn  between  the  axis  and  the 
diagonal  OS  may  be  used  for  the  focal  distance  of  a  point 
P  on  the  conic.  The  corresponding  directral  distance  of 
the  perpendicular  equals  the  directral  distance  of  the  point 
P  on  the  conic.  The  points  on  the  curve  are  located  by 
intersecting  the  measured  perpendicular  with  an  arc  centered 
at  the  focus  and  with  a  radius  equal  to  the  measured 
perpendicular. 


344 


PRACTICAL  MATHEMATICS 


The  point  of  intersection  of  any  curve  and  the  axis  is 
called  the  vertex. 


X2    i    Y2 

e<l       -fin  +-p— 1 


Fig.  133. — The  Construction  of  an  Ellipse  from  its  Analytic  Definition. 


The  parabola  has  a  single  vertex  and  a  single  focus.  Its 
vertex  lies  midway  between  the  focus  and  directrix. 

The  ellipse  has  two  vertices  V  and  V  and  two  foci 
F  and  F'   symmetrically  located  on  the  horizontal   axis. 


ECCENTRICITY  OF  CONICS 


345 


The  sum  of  the  radial  distances  from  the  foci  to  any  point 
on  the  curve  is  constant.     This  principle  is  applied  to  the 


e=l 

L 

/  ?/& 

''j/t 

/s     ' 

D 

~~~7&   / 

jri      ' 

/    i     / 

T 

S              /           / 

/         / 

/T 

/        / 

'9 

/      / 

/  J 

// 

/    p 

/    / 

■     o 

/       iv 

//^Focus 

w 

^  m  jl  m 

*\2  ^   2  "» 

\   H 

\  \ 

^  >> 

xl 

*£ 

\V 

1 

a 

vs. 

\ 


Fig.  134. — The  Analytic  Construction  of  the  Parabola  y  =2mX. 

trammel,  an  instrument  used  for  constructing  the  ellipse. 
There  are  two  equal  perpendiculars  which  can  be  intersected 


346 


PRACTICAL  MATHEMATICS 


by  the  arc  centered  at  F.  Therefore  an  hyperbola  has  two 
branches  which  lie  on  opposite  sides  of  the  directrix.  It 
has  two  vertices  V  and  V  and  two  foci  F  and  F'  symmetric- 
ally located  about  the  center  on  the  horizontal  axis. 


Fig.  135. — The  Analytic  Construction  of  the  Branches  of  the 
Hyperbola  from  its  Right  Focus. 


The  major  diameter  W  of  an  ellipse  or  hyperbola  is 
the  segment  of  the  horizontal  axis  lying  between  the  vertices. 
The  center  C  is  the  mid  point  of  the  major  diameter.  The 
minor  diameter  is  a  segment  of  the  perpendicular  bisector 


ECCENTRICITY  OF  CONICS 


347 


of  the  major  diameter.  The  semimajor  and  semiminor 
diameters  are  designated  by  a  and  b  respectively.  Table 
XXIII  gives  the  properties  of  conies. 

TABLE_XXIIL     PROPERTIES   OF  ELLIPSES   AND 
HYPERBOLAS 


Properties. 


Ellipse. 


1.  Value  of  e  in  terms  of  b  and  a 


*        a 


Hyperbola. 


=M 


2.  Value  of  b  in  terms  of  a  and  e 


ztaVl-e         ±aVe2-l 


3. 

Distance  of  extremity  of  minor 
from  focus 

axis 

V62+a2e2 

V&2+a2e2 

4 

Distance  of  focus  from  center .... 

±ae 

zhae 

5. 

Distance  from  focus  to  near  vertex . 

... 

i 

a(l-e) 

a(e-l) 

6 

Distance  from  focus  to  far  vertex. . 

a(l+e) 

a(l+e) 

7. 

Distance  from  directrix  to  center. . 

I 
... 

a 
e 

a 

e 

8. 

1 
Distance  from  directrix  to  near  vertex 

a  /i       \ 

>-d 

9.  Distance  from  directrix  to  far  vertex 


a 


d+e) 


U+e) 


10.  Distance  from  directrix  to  near  focus. .  i      —(1  — e2) 


(e2-D 


11. 

Distance  from  directrix  to  far  focus.  .  . 

^-d+e2) 

fa-M 

12. 

The    semi-parameter,    i.e.,    ordinate 
through  focus 

a(l-e2) 

a(e2-l) 

13. 

The  focal  distances  rx  and  r2  for  any 
point  on  the  curve 

T\  +r2  =  2a 

r\—  r2  =  2a 

CHAPTER  XVIII 
THE   FORMULATION   OF   GRAPHS 

1.  Writing  the  Equation  of  a  Curve.  When  the  varia- 
tion of  the  elements  of  a  graph  are  expressed  as  an  equation 
the  graph  is  said  to  be  formulated.  We  shall  assume  that 
our  familiarity  with  the  families  of  curves  plotted  in  the 
preceding  chapters  will  enable  us  to  recognize  a  parabola  or 
an  hyperbola. 

Many  experimental  curves  will  approximate  to  the 
graph  of  (1)  or  (2). 

(1)  y  =  bxm,  (2)  y  =  bx-m. 

This  assumption  implies  that  for  the  specific  curve  under 
consideration,  there  will  be  a  definite  value  for  b  and  a  definite 
value  for  m,  and  further  that  the  curve  (1)  passes  through 
the  origin  whereas  (2)  does  not. 

Select  any.  two  points  on  the  curve  and  measure  the 
pairs  of  coordinates  for  two  points,  abbreviate  these  by 
(x\,  yi)  and  (x2,  2/2).  If  the  curve  is  parabolic  each  pair 
of  coordinate  values  will  validate  (1),  i.e.,  the  equation  (1) 
will  hold  true  when  x\  and  yi,  and  also  when  X2  and  2/2  are 
substituted  therein.  We  obtain  equations  (a)  and  (6)  by 
substituting  the  coordinates  in  (1). 

(a)  2/i  =  toim,  (6)     y2  =  bx2m. 

(a)  and  (6)  are  simultaneous  equations  because  the  same 
constant  values  of  b  and  m  are  used  in  both  (a)  and  (b). 

348 


THE  FORMULATION  OF  GRAPHS       349 

We  can  eliminate  b  from  (a)  and  (b)  by  division  obtaining 
(c).     The  latter  can  be  transformed  for  m  by  logarization. 

»       E-S- ST- Div  (a)  by(6) 


(d)  log  —  =  m  log  — Log  of  a  power 

2/2  #2 


(e)  m  = — Div  (d)  by  coef  m 

X2 


(/)  .,     m  =  log;/i-log;/2  Log  of  quotient. 

W/  logXi  — logX2 


The  interpretation  of  •  (/)  states  that  the  exponent  of  the 
parabola  is  the  ratio  between  the  difference  of  the  logarithms 
of  the  ordinates  of  two  points,  to  the  difference  of  the  loga- 
rithms of  the  abscissas  of  the  same  two  points.  Logarizing 
(1)  we  obtain  (g), 

(g)  log  y  =  log  b+m  log  x Log  (1) 

(h)         .'.     log  6  =  log  y— m  log  x Trans  (g). 

The  interpretation  of  (h)  states  that  the  logarithm  of 
the  factor  b  may  be  obtained  by  subtracting  m  times  the 
logarithm  of  the  abscissa  of  any  point  from  the  logarithm 
of  the  ordinate  of  the  same  point. 

When  the  values  of  b  and  m  are  known  they  are  sub- 
stituted in  (1). 

If  the  graph  is  an  hyperbola  then  equation  (2)  is  applic- 
able. 


350  PRACTICAL  MATHEMATICS 

Ex.  1.  Show  that  the  values  of  m  and  b  for  an  hyperbola  may 
be  computed  from  (i)  and  ( j)  respectively. 

,«  _       logyi-logy2 

logZi-logx2 
(j)  log6=log2/+mlogx. 

Interpret  (i)  and  (j). 

Ex.  2.  Determine  the  formula  for  the  current-effort  curve 
EFGH  in  Fig.  136,  assuming  the  graph  is  a  parabola  passing  through 
the  origin.  Use  the  coordinates  of  the  points  F  and  H  in  (e) 
and  (h)  and  check  the  resulting  formula  for  the  point  G. 

Ex.  3.  Determine  the  formula  for  the  speed-current  curve 
ABCD  in  Fig.  136,  assuming  the  graph  is  an  hyperbola.  Use  the 
coordinates  of  the  points  A  and  D  in  (i)  and  (j)  and  check  the 
resulting  formula  for  the  point  C. 

2.  Equations  (1)  and  (2)  may  be  written  in  the  linear 
forms  (3)  and  (4)  by  logarization.  (3)  and  (4)  are  identical 
in  form  with  (5)  the  equation  of  a  straight  line. 

(3)  log  y  =  log  b+m  log  x. 

(4)  log  y  =  log  b  —  m  log  x. 

(5)  •   Y  =  A+BX. 

(3)  and  (4)  contain  the  variable  log  y  which  is  replaced 
by  Y  in  (5),  and  the  variable  log  x  which  is  replaced  by 
X  in  (5) .  Log  b  is  a  constant  and  is  replaced  by  the  constant 
A  in  (5)  and  m  or  —  m  are  constants  also  and  are  replaced 
by  B  in  (5). 

If  we  plot  (5)  we  will  obtain  a  linear  graph.  Y  is  the 
ordinate  and  X  the  corresponding  abscissa  for  any  point 
on  the  line.  A  is  the  intercept  on  the  Y  axis  and  B  is  the 
slope  of  the  line. 

To  plot  Y  means  to  plot  logarithms  of  y  vertically  and 
to  plot  X  means  to  plot  logarithms  of  x  horizontally. 

3.  The  necessity  of  determining  the  numeric  values 
of  log  y  and  log  x  for  plotting  is  obviated  by  the  use  of 


THE  FOEMULATION  OF  GRAPHS 


351 


logarithmic  cross-section  paper.  Such  paper  is  shown  in 
Fig.  137,  which  represents  two  independent  unit  logarithmic 
squares  called  log-units  arranged  side  by  side  for  convenience. 
Figs'.  139  and  140  represent  two  unit  logarithmic  squares 
arranged  vertically  one  above  the  other.  A  number  of 
log-units  may  be  united  for  extended  plotting. 

A  log-unit  is  numbered  on  its  axes  from  1  to  10  in  sequence. 
The  spacing  of  log  paper  is  unlike  the  uniform  spacing  of 
squared  cross-section  paper  but  follows  the  spacing  of  a 
log  scale  of  a  slide  rule.     The  intersection  of  the  axes,  i.e., 


10      iO     dO      70     80     SO    100 

Amperes 
Fig.  136. — The  Extrapolation  of  Curves  from  Logarithmic  Plotting. 

the  zero  point  for  uniform  scales,  is  marked  1  for  log  scales 
because  log  1=0.  The  log-unit  is  one  unit  in  length 
and  therefore  the  extreme  ends  of  the  axes  are  marked  10 
since  log  10=1.  The  divisions  of  the  log  paper  are  man- 
tissas and  therefore  the  scale  may  be  multiplied  by  a  power 
of  ten. to  accommodate  a  range  of  numbers  from  1-10, 
10-100,  100-1000  or  .1-1,  .01-.1,  .001-.01,  etc.  Joining  two 
log-units  increases  the  range  of  the  combined  scale  from 
1-100  or  any  equivalent. 

4.  The  points  on  the  graphs  in  Fig.  136  are  replotted  in 
the  two  log-units  of  Fig.  137.  For  every  point  on  the  squared 
paper,  Fig.  136,  there  is  a  corresponding  point  on  the  log 


352 


PRACTICAL  MATHEMATICS 


paper,  Fig.  137.  These  corresponding  points  are  marked 
with  like  letters  for  ready  identification.  By  extending 
the  linear  graphs  of  Fig.  137,  additional  points  such  as  L, 
K  and  J  may  be  obtained  and  thereby  extrapolate  the  orig- 
inal data.  Any  point  such  as  (15,  20)  of  the  squared  paper 
will  have  the  same  coordinates  (15,  20)  on  the  log  paper. 

Ex.  4.  Measure  the  slope  and  intercepts  of  ABCD  in  Fig.  137 
and  compare  their  values  with  the  results  of  Ex.  2. 

Ex.  5.  Measure  the  slope  and  intercepts  of  EFGH  in  Fig.  137 
and  compare  their  values  with  the  results  of  Ex.  3. 


W 

u 

2.1 


Amperes  10-100 

2  3         A567891 


Amperes  10-100 

2  3         4       8     6    7 


SPEED-CURRENT  CURVE 


D.C.  SERIES  MOTOR 


10 
9 
8 
7 

5  ij? 


Fig.  137. — Plotting  on  Logarithmic  Paper. 


Observation.  When  parabolic  and  hyperbolic  graphs  are 
plotted  on  logarithmic  paper  the  slope  of  the  linear  graph  is 
the  exponent  of  the  horizontally  plotted  variable  and  the  inter- 
cepts on  the  vertical  axis  is  the  coefficient  of  the  same  variable. 

5.  In  Fig.  138,  (a)  is  the  standard  parabola  y  =  x15.  (c) 
and  (d)  illustrate  the  effect  of  changing  the  coefficient, 
(e)  is  the  standard  hyperpola  y  =  x~15  which  illustrates  the 
effect  of  changing  the  sign  of  the  exponent.  Figs.  139  and 
140  represent  (a),  (6),  (c)  and  (d)  plotted  upon  logarithmic 
paper.  The  log-unit  ABUV  corresponds  to  the  small 
square  AB  UV  of  Fig.  138.  The  log-unit  BCW U  corresponds 
to  the  rectangle  BCWU  in  Fig.  138.    The  log-unit  WDFU 


THE  FORMULATION  OF  GRAPHS 


353 


corresponds  to  the  large  square  WDFU  in  Fig.  138.  The  log- 
unit  UFEV  corresponds  to  the  rectangle  UFEV  in  Fig.  138. 
The  logarithmic  paper  provides  a  means  for  magnifying 
the  region  close  to  the  origin.  The  edges  of  the  logarithmic 
paper  are  subdivided  into  tenths  of  the  principal  divisions. 


0     N    1  IV 


Fig.  138.— Plotting  of  y  =  bx*m  on  Squared  Paper. 


A  scale  of  equal  parts  indicates  the  manner  of  graduating 
the  principal  divisions  and  the  subdivisions  of  the  log  paper. 
The  graphs  (a),  (c)  and  (d)  in  Figs.  139  and  140  are 
parallel  and  their  constant  slope  is  measured  by  the  tangent 
of  the  angle  of  inclination.  The  angle  is  56°  20'.  The 
tan  56°  20' =  1.5,  which  corresponds  to  the  exponent  of  the 
variable  x  for  the  three  equations  (a),  (6)  and  (d).     The 


354 


PRACTICAL  MATHEMATICS 


9  10 


2  2.5       3      N        4  5         6         7      8      9     10 

LogX 
Fig.  139. — Logarithmic  Plotting  of  y  =  bx*m. 


THE  FORMULATION  OF  GRAPHS 


355 


Scale  of  Equal  Parts 

10    11    12     13    14    15     16    17     18     19    20 


Log 
1     Scale 


M     2 3 4     5    G   7 

IHIIIIIIII  I  I  I  MMi|]Mlllllll|lllllllll|llll 


Fig.  140. — Logarithmic  Plotting  of  y  — 


356  PRACTICAL  MATHEMATICS 

slope  of  (e)  in  Fig.  140  is  negative.  The  angle  is  123°  40'. 
The  tan  123°  40' =  —1.5  which  corresponds  to  the  exponent 
of  the  variable  x  in  (e).  The  vertical  intercept  of  (c)  is 
at  Pi,  indicating  2  which  is  the  coefficient  of  the  variable 
x  in  (c).  The  vertical  intercept  of  (d)  is  at  R7  which  is 
obtained  by  extending  MT  through  the  next  log-unit.  R7 
indicates  .5  which  is  the  coefficient  of  the  variable  £  in  (d). 
The  vertical  intercept  of  (e)  is  at  U  indicating  1  which  is 
the  coefficient  of  the  standard  hyperbola  (c). 

The  graph  (a)  should  have  been  represented  in  the  log 
unit  WUFD  but  in  order  to  avoid  over-crowding  the  same 
purpose  is  served  by  representing  it  in  log-unit  CBUW 
providing  we  retain  the  proper  range  of  values  of  the  scale 
divisions. 

By  extending  the  graphs  into  the  next  log-unit  a 
greater  range  of  values  of  the  variables  may  be  obtained. 
This  extension  may  be  provided  for  in  a  better  way  so  as  to 
economize  in  paper  and  space  and  at  the  same  time  not 
sacrifice  any  accuracy.  The  extension  of  MT  gives  the 
segments  MR7  and  R7N.  Locate  M'  vertically  above  M 
and  draw  M'Rj  parallel  to  MT.  Then  M'R7  represents 
MR7  for  which  the  vertical  scale  should  be  read  in  tenths. 
Locate  R7"  on  the  same  horizontal  line  with  R7' '.  Through 
R7"  draw  R7"N'  parallel  to  MT.  Then  RJ'N'  represents 
R7N  for  which  the  horizontal  scale  should  be  read  in  tenths. 
Vertically  under  T  locate  V '.  Draw  T'Rg  parallel  to  MT. 
Then  T'Rq  represents  the  extension  of  MT  into  the  next 
higher  log-unit.  By  continuing  this  process  the  entire  graph 
may  be  represented  in  a  single  log-unit. 

Ex.  6.  Plot  upon  one  log-unit  the  graphs  for  reading  the 
squares  and  cubes  of  numbers  and  the  reciprocals  of  numbers. 
Show  how  these  graphs  may  be  used  for  obtaining  square  roots 
and  cube  roots. 

Observation.  Logarithmic  paper  gives  a  ready  means  for 
determining  the  constants  of  a  graph  of  the  type  y  =  bx±m.    It 


THE  FORMULATION  OF  GRAPHS  357 

serves  as  a  quick  method  for  plotting  such  a  curve  since  it  requires 
a  minimum  amount  of  data.  It  magnifies  any  desired  region 
of  uniformly  divided  cross-section  paper.  An  entire  graph 
may  be  represented  in  one  log-unit. 

6.  If  a  given  curve  is  not  of  the  type  y  =  bxm  the  graph 
will  not  be  linear  upon  logarithmic  paper.  The  equation 
y  =  bx±m  is  a  close  approximation  if  a  straight  line  may  be 
drawn  through  the  average  of  a  number  of  points  plotted 
on  logarithmic  paper. 

If  very  great  accuracy  is  sought  in  determining  the 
formula  for  a  very  precise  determination  we  may  use  the 
power  series  (25),  (26),  (27)  or  (28),  suggested  in  Chap. 
XVI.  A  definite  number  of  terms  usually  not  to  exceed 
four,  is  chosen  for  the  right-hand  member.  This  form 
requires  us  to  determine  one  constant  for  each  term.  Four 
simultaneous  equations  are  written  containing  the  four 
pairs  of  measured  coordinates  taken  from  the  curve.  The 
solution  of  four  equations  gives  the  numeric  values  of  the 
constants. 

We  are  often  guided  in  the  choice  of  a  suitable  formula 
by  the  shape  of  the  curve.  If  one  of  the  variables  becomes 
indefinitely  large  while  the  other  variable  approaches  a 
limiting  value  then  use  (6), 

fa\  ax 

(6)  y  = 


-b' 

Other  equations  which  are  suggested  for  the  formulation 
of  graphs  are  given  in  (7),  (8)  and  (9): 


(7)  *=i+3* 

(8)  y  =  (a+bx)», 

(9)  y  =  a+bxn. 

Ex.  7.  Plot  (6),  (7),  (8),  and  (9),  when  a  =2,  b  =  1,  e  =  .5. 

Ex.  8.  Plot  (6),  (7),  (8),  and  (9),  when  a  =2,  b  =1,  e  =  1. 


358  PRACTICAL  MATHEMATICS 

Ex.  9.     Plot  (6),  (7),  (8),  and  (9),  when  a  =2,  b  =2,  e  =  l, 
Ex.  10,  Plot  (6),  (7),  (8),  and  (9),  when  a  =  l,  6  =  1,  e=l. 

7.  Any  equation  which  may  be  restated  in  the  linear 
form  (5)  may  be  plotted  as  a  linear  graph, 

(5)  Y=A+BX. 

Consider  the  logarithmic  curves  represented  in  Fig.  113, 
their  equations  are  (10)  and  (11): 

(10)  y  =  \ogx; 

(11)  y  =  \ogex; 

(12)  Y  =  X. 

(.10)  and  (11)  are  forms  of  the  linear  equation  (12)  in 
which  Y  represents  y,  and  X  represents  log  x.  If  y  be  plotted 
vertically  to  a  uniform  scale  and  x  horizontally  to  a  loga- 
rithmic scale  the  graph  of  (10)  will  be  a  straight  line  with 
a  slope  of  unity.  Paper  prepared  in  this  manner  is  called 
semi-logarithmic  paper.  By  means  of  the  scale  of  equal 
parts,  the  logarithmic  paper  of  Figs.  139  and  140  may  be 
used  as  semi-logarithmic  paper. 

(6)  may  be  transformed  to  read  (13)  which  corresponds 
to  the  intercepts  form  (14)  of  (5). 

y. 


(14)  f+li-1. 

B 

y 

Therefore  if  y  is  plotted  vertically  and  -  is  plotted 

x 

horizontally  both  being  uniformly  scaled  then  the  graph 

is  linear.   Fig.  141  shows  (6)  plotted  in  the  usual  way  as  an 


THE  FORMULATION  OF  GRAPHS 


359 


hyperbola  II  and  also  (13)  the  equivalent  of  (6)  plotted  in 
the  linear  form  I.     The  vertical  and  horizontal  intercepts 

of  I  are  shown  as  a  and  ——respectively,  which  verify  (13). 

o 


11 

10 

I 

n    y    (  1    IS  PLOTTED  TO    Y     AND  I   SCALES 

Y=  — ■!                                       x           1 

X-6   |  U    ,S  PLOTTED  TO  Y   AND   X  SCALES 

9 
8 

\ 

\ 

I 

ill 

7 

Ye 

5 

I 

f 

3 
t 

1 

) 

t 

i 

v 

L 

X 

-1      [  0 


Fig.  141. — Ordinary  and  Linear  Representation  of  y  = 


x-b 


The  vertical  line  which  is  located  one  unit's  distance  to 
the  right  of  the  y  axis  is  the  axis  of  x,  and  is  called  the 
x  index.     A  straightedge  which  joins  the  origin  with  any 


360 


PRACTICAL  MATHEMATICS 


point  on  the  linear  graph  I  cuts  the  index  at  a  point  which 
indicates  the  corresponding  value  of  x.  The  y  axis  and  x 
index  are  graduated  alike.  The  latter  property  serves  as 
a  method  for  the  rapid  plotting  of  the  linear  represen- 
tation of  (6). 

(7)  may  be  transformed  to  read  (15)  which  corresponds 
to  the  intercepts  form  (14)  of  (5). 


* 4 


Fig.  142. — Ordinary  and  Linear  Representation  of  y  = 


1+ex2' 


(15) 
(14) 


£+1=1, 


a 

c 


A^-A     i* 
B 


Therefore  if  y  is  plotted  vertically  and  —  is  plotted 

x 

horizontally  both  being  uniformly  scaled  then  the  graph 


THE  FORMULATION  OF  GRAPHS       361 

is  linear.    Fig.  142  shows  (7)  plotted  in  the  usual  way  as  a 
cubic  hyperbola  II  and  also  in  the  linear  form  I. 

The   vertical   and   horizontal   intercepts  are   shown   as 

a 

—  and  a  respectively.     The  vertical   line  which   is   located 
e 

one  unit's  distance  to  the  right  of  the  y  axis  is  the  x  index. 

The  x  index  should  be  graduated  as  square   roots  of   the 

paralleled  numbers   on    the  y  axis.     A  straightedge  which 

joins  the  origin  with  any  point  on  the  linear  graph  I  cuts 

the  index   at   a   point  which  indicates    the  corresponding 

value  of  x.     The  latter  property  serves   as  a  method  for 

the  rapid  plotting  of  the  linear  representation  of  (7). 

Ex.  11.  Show  that  the  conic  equations  (37)  and  (42),  Chapter 
XVI,  may  be  plotted  as  linear  graphs  on  cross-section  paper  in 
which  the  vertical  and  horizontal  rulings  are  graduated  in  squares 
of  numbers  similar  to  the  x  index  in  the  preceding  paragraph. 

8.  The  sine  curve,  Fig.  143,  is  expressed  as  equation  (16) 
and  is  identified  with  the  linear  form  (12). 

(16)  y  =  smx.  N 

(12)  Y  =  X. 

In  Fig.  143,  OABCDE  represents  half  of  the  sine  loop 
plotted  in  the  usual  way.  The  vertical  and  horizontal  scales 
are  uniform  scales  of  numbers  and  angles  respectively. 
If  the  horizontal  scale  be  made  to  correspond  to  sines  of  x, 
as  shown  in  the  sinusoidal  scale  of  angles,  then  the  half  loop 
becomes  a  linear  graph  OE.  Like  letters  in  each  graph 
represent  corresponding  points. 

Ex.  12.  Prepare  a  table  of  equations  both  algebraic  and  non- 
algebraic  which  are  representable  by  linear  graphs  when  plotted 
to  specially  scaled  cross-section  paper.  In  parallel  columns  indicate 
the  nature  of  the  scales  for  the  two  axes. 

9.  Exponential  Equations.  When  one  of  the  variables 
in  an  equation  appears  as  an  exponent  of  a  constant  base 


362 


PRACTICAL  MATHEMATICS 


§  *-M-u       J 


THE  FORMULATION  OF  GRAPHS  363 

the  equation  is  called  an  exponential  as  illustrated  in  (17), 
(18),  (19)  and  (20). 

(17)  y  =  ax,  (18)  y  =  bacx. 

(19)  y=skx,  (20)  y-d  =  bBkix-m\ 

Usually  the  constant  base  is  the  Napierian  number  (e). 
(17)  and  (18)  may  be  transformed  into  (24)  and  (28)  re- 
spectively, by  the  introduction  of  a  constant  multiplier  for 
the  variable  exponent. 

(21)  log£i/  =  zlogea;  (25)    logs  y  =  \oge  6+ ex  log£  a; 

(22)  loge  a  =  a  constant  K ;  (26)  c  log£  a  =  n,  log£  b  =  B ; 

(23)  /.   \ogsy  =  Kx;  (27)    \ogsy  =  B+nx; 

(24)  /.  y=sKx;  (28)  y  =  eB+nx. 

There  are  many  applications  of  this  law.  It  has  been 
styled  the  Compound  Interest  Law  by  Lord  Kelvin. 

Plot  the  following  equations  using  semi-log  paper  and 
consult  the  exponential  table  XXIII. 

Ex.  13A.  (24)  y=zkx.  Use  the  following  values  of  k,  .1,  .5, 
1,  2,  5. 

Ex.  13B.  (29)  y  =z~kx.  Use  the  following  values  of  k,  .1,  .5, 
1,  2,  5. 

Ex.  14.  The  friction  of  a  rope  or  belt  on  a  pulley  or  cylinder 
is  expressed  by  (30).  Plot  (30)  in  which  T2  is  the  maximum 
tension  and  7\  the  minimum  tension,  a  the  angle  of  contact,  and 
£jl  the  coefficient  of  friction,  [i  for  hemp  rope  on  cast-iron  is  .2-.4, 
and  for  wire  rope  .5. 

(30)  fr£±Ma 

T 

make  —  the   independent   variable   and   plot    a   from  0   to  180° 

1 1 
when  \l  m  .2,  .25,  .3,  .35,  .4,  .45,  5. 

Ex.  15.  The  necessary  distance  (d)  for  the  separation  of 
conductors  to  avoid  corona  is  given  in  (31)  in  which  D  is  the 
diameter  in  inches,  d  is  the  distance  in  inches,  E  is  effective  kilo- 


364 


PRACTICAL  MATHEMATICS 


volts  between  conductors  and  K  is  a  constant.     Assume  i57  =  150 
and  A' =39.9. 


(31) 

d=- 

D    KD-S 

I?        ' 

Ex.  16.    Variation  of  voltage  with  self  induction. 

Rt 

(32)                                E=RI-Rhz   L. 

1.6 

1.4 

\ 

1.2' 
1.0 
.8 
.6 
.4 
.2 

\ 

e=  Logarithmic  Base  =  2.7183 

1  Y  =  e-x+.5e-5a; 

2  Y  =  e-a"+.2€-2X 

3  Y=e-x+.ie-,a; 

4  Y=e-* 

5  Y  =  e-a;-.i6-3a; 

6  Y  =  e-x-.2e-3a; 

7  Y  =  e  -x-l.l€-2X 

8  Y  =  e-x-.5e-9iK 

\  w 

H^ 

L 

^o 

/ 
/ 

^^S^J 

r 

I 
l 
l 

\ 1 

^^ 

^^ 

»_ 

\s^~ 

?!-=a*&«j 

Y  „ 

/ 

b=o^s 

/ 1 
i  j 

2      .4       .< 

5        .8 

0      1 

2      1.4 

6      lU 

0      2 

2      2 

4      2 

6 

-.2 
-.4 

/  ; 

X- 

/    i 
7  ' 

-.6 

-.8 

i8 

-1  ft 

L?   ' 

Fig.  144. — The  Exponential  Family. 

Jo  is  the  value  of  the  current  when  t  =  0.  See  Figs.  191 
and  192. 

10.  The  standard  form  of  the  exponential  may  be 
written  (33)  or  (34),  but  the  exponent  appears  negative  for 
a  majority  of  technical  applications. 


(33) 


¥-*", 


(34) 


y=e 


THE  FORMULATION  OF  GRAPHS 


365 


(34)  is  illustrated  as  curve  4  in  Fig.  144.  It  crosses  the 
Y  axis  at  the  point  1.  To  the  left  of  the  Y  axis  it  rises 
rapidly  as  x  becomes  greater  negatively.  On  the  right  of 
the  Y  axis  it  falls  less  rapidly  as  X  increases  positively. 
The  plotting  of  (34)  on  semi-logarithmic  paper  determines 
a  straight  line  with  a  negative  slope  (  —  1). 

Many  exponential  equations  are  combinations  of  two 
exponential  forms  as  represented  by  (35)  and  as  illustrated 


Fig.  145. — The  Exponential  Family. 


in  Fig.  144.  Curves  2  and  7  may  be  compared  to  show  the 
effect  upon  the  standard  when  a  second  exponential  is 
either  added  or  subtracted.  The  other  curves  serve  to 
show  the  effect  of  changing  the  constants  in  (35). 


(35) 


y=z     ±az 


-bx 


Other  exponential  curves  of  the  type  (35)  are  shown  in 
Fig.  145,  which  have  been  plotted  separately  for  clarity. 


366 


PKACTICAL  MATHEMATICS 


Ex.  17.  Plot  y  =  i{ex+e~x};  this  particular  exponential  is 
designated  an  hyperbolic  cosine  and  is  abbreviated  cosh  x. 
Assume  values  of  x  between  0  and  2. 

Ex.  18.  Plot  y=h{zr—  £~x] I  this  particular  exponential  is 
designated  an  hyperbolic  sine  and  is  abbreviated  sinh  x.  Assume 
values  of  x  between  0  and  2. 

Ex.  19.  Plot  the  following  exponentials  in  Table  XXIII  on 
semi-logarithmic  paper  and  extend  the  range  of  the  tables  by 
adding  parallels  to  the  linear  graphs,  as  described  in  paragraph 
5,  page  356.     Add  the  supplementary  scales  for  each  parallel  line. 


TABLE  XXIII.     EXPONENTIALS 


ex 

Exponent  x. 

e~x 

ex 

Exponent  x. 

e~x 

1.000 

.0 

1.000 

3.669 

1.3 

.273 

1.105 

.1 

.905 

4.055 

1.4 

.247 

1.221 

.2 

.819 

4.482 

1.5 

.223 

1.350 

.3 

.741 

4.953 

1.6 

.202 

1.492 

.4 

.670 

5.474 

1.7 

.183 

1.649 

.5 

.607 

6.050 

1.8 

.165 

1.822 

.6 

.549 

6.686 

1.9 

.150 

2.014 

.7 

.497 

7.389 

2.0 

.135 

2.226 

.8 

.449 

8.166 

2.1 

.122 

2.460 

.9 

.407 

9.025 

2.2 

.111 

2.718 

1.0 

.368 

9.974 

2.3 

.100 

3.004 

1.1 

.333 

11.023 

2.4 

.091 

3.320 

1.2 

.301 

12.182 

2.5 

.082 

Ex.  20.  The  exponential  curves  shown  in  Figs.  144  and  145  are 
obtained  by  addition.  If  the  component  exponential  curves  are 
replotted  on  semi- logarithmic  paper  can  they  be  added  to  produce 
the  resulting  curves  shown  in  Figs.  144  and  145? 

Ex.  21.  Show  how  an  exponential  table  may  be  constructed 
from  a  table  of  hyperbolic  logarithms  or  from  a  table  of  hyperbolic 
sines  and  cosines. 


CHAPTER  XIX 


THE   USE   OF   POLAR   PAPER 


1.  In  the  preceding  chapters  it  was  shown  that  a  point 
was  completely  determined  by  its  coordinates,  i.e.,  its  dis- 
tances from  two  fixed  lines  called  axes.  The  axes  may  be 
represented  in  a  rectangular  or  oblique  position  and  the 
scales  thereon  may  be  graduated  uniformly  or  nonuniformly. 

In  Fig.  146,  the  horizontal  axal  line  has  a  fixed  left-hand 
extremity  called  a  pole  or  origin.     The  point  Pi  may  be 


X2  Xy 

Fig.  146. — Polar  Coordinates. 

located  by  the  pair  of  rectangular  coordinates  x\  and  y\y 
and  it  may  be  located  also  by  a  pair  of  polar  coordinates 
Ri  and  0i.  The  radius  vector  (R)  is  the  distance  measured 
radically  from  the  pole  to  the  point.  The  vectorial  angle 
(6)  is  the  angle  formed  by  the  radius  vector  and  the  horizon- 
tal axal  line.  Likewise  the  point  P2  may  be  designated  and 
located  by  its  rectangular  coordinates  £2,  2/2  and  by  its 
polar  coordinates  R2,  62. 

A  curve  may  have  its  formula  expressed  in  terms  of  the 
variables  R  and  8  as  well  as  in  terms  of  the  variables  x  and  y. 
The  relations  between  R,  0,  x  and  y  are  expressed  in  (1), 
(2),  (3)  and  (4). 

367 


368  PRACTICAL  MATHEMATICS 

(1)  x  =  RcosQ. 

(2)  y  =  R  sin0. 


(3)  R  =  ±Vx2+y2. 

y 


(4)  tanG 


x 


Therefore  an  equation  in  x  and  y  may  be  rewritten  in 
terms  of  R  and  0  by  substituting  from  (1)  and  (2).  Thus 
(5)  becomes  (6)  and  reduces  to  (7) . 

(5)  y  =  bxm. 

(6)  #sin0  =  6iTcosw0. 
„  sec  0 


(7)  H  = 

V  b  cot  0 

2.  In  order  to  obviate  the  necessity  for  measuring  the 
angles  and  radii  vectors  for  polar  plotting  we  may  use 
polar  coordinate  paper,  which  is  prepared  for  the  trade. 
Polar  paper  as  shown  in  Fig.  147,  is  divided  by  concentric 
circles  and  radial  lines.  When  the  circles  are  equally 
spaced  the  radii  are  uniformly  divided.  When  the  radii 
are  equally  spaced  the  circles  are  uniformly  divided.  The 
center  of  the  circles  is  the  pole.  The  horizontal  radius 
extending  to  the  right  of  the  pole  is  the  axal  line  from  which 
positive  and  negative  angles  are  read  in  counter-clockwise 
and  clockwise  direction  respectively.  The  outer  circumfer- 
ence is  graduated  in  degrees  and  %  measure.  Other  points 
of  divisions  may  be  supplemented  to  indicate  positive  and 
negative  multiples  and  submultiples  of  radians. 

3.  To  use  polar  coordinate  paper  first  solve  the  polar 


THE  USE  OF  POLAR  PAPER 


369 


formula  for  R  and  prepare  a  table  of  values  for  R  and  0. 
Designate  the  radian  scale  so  as  to  comprehend  the  greatest 
value  of  R  in  the  table  and  indicate  the  first  six  positive  and 
negative  radians.  If  6  is  expressed  in  degree  measure 
then  its  unit  value  or  radian  measure  may  be  obtained  from 
Table  VIII.     Although  the  angles  may  be  plotted  in  degrees 


Fig.  147. — The  Plotting  of  Polar  Equations. 

the  solution  of  R  can  be  obtained  usually  only  after  8  is 
expressed  in  unit  measure. 


Ex.  1.     Plot  (8)  R  =  6.     The  values  of  G  and  R  are  given  in 
Table  XXIV.    The  corresponding  curve  is  shown  as  I  in  Fig.  147. 


370 


PRACTICAL  MATHEMATICS 


There  are  two  circles  marking  20  unit  divisions  on  each  radius. 
Curve  I  recedes  radially  6.28  units  in  one  complete  convolution. 

One  complete  convolution  corresponds  to  a  rotation  of  the 
radius  vector  through  360°,  or  2%  radians.  I  is  known  as  a 
standard  linear  spiral,  or  the  spiral  of  Archimedes. 


TABLE  XXIV.    THE  STANDARD  LINEAR  SPIRAL. 


R 

0 

.262 

.524 

1.05 

1.57 

2.09 

2.62 

3.14 

3.66 

4.19 

4.71 

5.23 

5.77 

Degrees 

0° 

15° 

30° 

60° 

90° 

120° 

150° 

180° 

210° 

240° 

270° 

300° 

330° 

Radians 

0 

.262 

7T 
12 

.524 

1.05 

1.57 

2.09 

2.62 

3.14 

3.66 

4.19 

4.71 

5.23 

5.77 

llir 
6 

x  meas- 
ure 

0 

"6 

~3 

IT 

~2 

2t 
3 

5* 
6 

T 

7* 
6 

4t 
3 

3* 
2 

5x 
3 

6.28 
360° 
6.28 

2*- 


4.  The  graph  of  (9)  R  =  2Q  is  shown  as  curve  III  in  Fig. 
147.  The  radial  ordinates  in  (9)  are  correspondingly  twice 
the  magnitude  of  the  radial  ordinates  in  (8). 

The  graph  of  (10)  R  =  .50  is  shown  as  curve  II  in  Fig.  147. 
The  radial  ordinates  of  (10)  are  correspondingly  one-half 
the  magnitude  of  the  radial  ordinates  in  (8). 

Observation.  A  spiral  is  the  polar  graph  of  a  polar  equa- 
tion which  does  not  contain  trigonometric  functions.  The 
standard  linear  spiral  R  =  Q  is  the  polar  representation  of  the 
rectangular  linear  graph  y  =  x.  The  radial  recession  or  depar- 
ture of  the  linear  spiral  is  proportional  to  its  angular  advance. 
The  numeric  coefficient  of  0  increases  the  departure  propor- 
tionately. When  the  absolute  or  constant  term  is  missing 
from  the  polar  equation  the  spiral  passes  through  the  origin. 


Ex.  2.     Plot  (8),  (9)  and  (10)  for  negative  values  of  6  and  R. 

Negative  angles  are  laid  off  clockwise  in  accordance  with  the 
convention  of  signs  in  trigonometry.  Negative  values  of  R  are 
to  be  plotted  in  a  contrary  sense  to  positive  values  and  may 
be  laid  off  by  extending  the  radial  lines  back  through  the 
pole.  The  positive  and  negative  halves  of  spirals  represent 
cams  in  machine  design. 


THE  USE  OF  POLAR  PAPER  371 

Plot  the  following  spirals  for  positive  and  negative  values  of  6. 

Ex.3.  (11)  #  =  82,  (12)  #=282,  (13)  #  =  .582. 

Ex.4.  (14)  #  =  83,  (15)  R=2V,  (16)  R=.UK 

Ex.5.  (17)  #  =  8-',  (18)  #=58-\  (19)  R=2oQ~\ 

Ex.6.  (20)  22  =  1+6,  (21)  #=2+.58,  (22)  #=28  +  .582. 

Spirals  (8)  to  (16)  are  parabolic  forms,  whereas  spirals  (17)  to 
(19)  are  hyperbolic  forms.  They  are  so  named  on  account  of  the 
resemblance  to  similar  rectangular  equations. 

Ex.  7.  Place  a  sheet  of  transparent  polar  paper  upon  a  sheet 
of  squared  paper  so  that  the  respective  pole  and  origin  are  coincident 
and  so  that  the  respective  axal  line  and  x  axis  are  coincident.  A 
straight  line  which  passes  through  the  pole  and  origin  will  be  ex- 
pressed as  (23)  in  polar  coordinates  and  as  (24)  in  rectangular 
coordinates. 

(23)  tan  8=6,  (24)  y  =bx. 

Any  linear  graph  which  cuts  the  pole  will  have  a  constant  slope 
as  shown  in  (23). 

Ex.  8.    Show  that  the  polar  form  of  (25)  is  (26). 

(25)  y=a+bx.  (26)  fl(sin  6-6  cos  8)  =a, 

Ex.  9.    Show  that  the  polar  form  of  (27)  becomes  (7). 

(27)  log  y=\ogb+m  logx. 

Plot  the  following  equations: 
Ex.10.     (28)        #=sin8,      (29)        R  =  cos  8, 
Ex.11.     (30)        fl=csc8.      (31)         fl=sec8. 

5.  The  conies  may  be  constructed  by  means  of  polar 
coordinates  from  (32).  In  such  cases  the  focus  of  the 
conic  is  located  at  the  pole. 

(32)  R 


l—e  cos 


372  PRACTICAL  MATHEMATICS 

In  (32)  e  is  the  eccentricity  of  the  conic  and  I  is  the 
ordinate  above  the  focus. 

Ex.  12.    Plot  (32)  when  1  =  1  and  e  =  .5,  1  and  1.5. 
6.  The  polar  equation  of  a  circle  is  given  in  (33). 

(33)  a2  =  R2+r2-2Rr  cos  (6 -a). 

(33)  is  identified  with  Fig.  148,  in  which  a  =  radius  of 
the  circle,  R  =  the  polar  ordinate  of  any  point  on  the  cir- 
cumference and  6  the  corresponding  angular  ordinate; 
r  =  the  polar  ordinate  of  the  center  and  a  the  corresponding 
angular  ordinate.  From  the  figure  it  is  seen  that  for  a 
given  value  of  0  there  will  be  two  values  of  R  corresponding 
to  the  points  Pi  and  P2,  respectively. 

Ex.  13.  Show  that  (33)  reduces  to  (34)  when  the  pole  is  coin- 
cident with  the  lower  extremity  of  a  vertical  diameter. 

(34)  fl=2rsin6. 


Fig.  148.— The  Polar  Circle. 

Ex.  14.  Show  that  (34)  reduces  to  (35)  when  the  center  of  the 
circle  coincides  with  the  pole.    What  is  the  interpretation  of  (35). 

(35)  R  =  a. 

7.  The  Logarithmic  Spiral.  The  logarithmic  spiral  is 
also  called  the  equiangular  spiral  and  represents  an  expo- 
nential curve  and  corresponds  to  4  in  Fig.  144.  Its  equation 
is  given  in  (36)  and  (37).  It  takes  its  name  from  (37). 
It  is  a  curve  which  has  considerable  application  in  the  design 


THE  USE  OF  POLAR  PAPER 


373 


of  milling  cutters  owing  to  the  fact  that  the  curve  cuts 
each  radius  vector  with  a  constant  angle.  The  angle  shown 
in  Fig.  149  is  124°  42'.  The  tangent  of  124°  42' =  1.44  which 
is  also  the  modulus  of  the  system  of  logarithms  (6  =  2)  used 
in  plotting  Fig.  149. 


(36)     R  =  b\ 


(37)  log6#  =  0. 


(38)     R  =  z 


K8 


Eq.  (36)   is   a  special  form  of   (38)  in  which  K  is    an 
arbitrary  constant.     When  the  Napierian   base  is  used  we 


Fig.  149. — The  Logarithmic  or  Equiangular  Spiral. 

have   the   portrayal   of   the   exponential   curves   shown   in 
Fig.  144. 

A  table  of  values  between  0,  K%,  log  R  and  R  is  prepared 
for  the  special  value  6  =  2.  The  curve  is  plotted  between 
R  and  0.  Any  radius  vector,  which  bisects  the  angle  formed 
by  its  two  adjacent  radii  vectors,  is  a  mean  proportional 
between  them.     In   Fig.    149,   the   logarithmic   spiral   has 


374  PRACTICAL  MATHEMATICS 

been  plotted  with  radii  vectors  displaced  by  one-half  of  a 
radian.  Thus  LO  is  a  mean  proportional  between  OK  and 
OM.  Noting  the  radian  measure  of  the  respective  angles 
for  the  three  mentioned  radii  we  have 

(a)       OK  =  bA5,       (b)         OL  =  b5,         (c)        OM  =  b55. 


01?    b 


10 


»  •'•    °*=b4-5=oS=^ 


What  is  the  character  of  that  part  of  the  curve  where 
0  is  negative. 

Ex.  15.  Construct  equiangular  spirals  for  bases  2,  2.25,  2.5, 
3;  also  a  spiral  to  have  its  slope  =  .434  when  measured  to  any 
radius  vector. 

8.  Polar  coordinates  are  commonly  used  to  plot  the 
measurements  of  the  intensity  of  illumination.  Another 
form  of  polar  paper  is  that  represented  in  the  familiar  discs 
or  record  cards  of  recording  instruments  of  the  Bristol  type. 
The  radial  lines  of  regular  polar  paper  are  replaced  by 
semi-radial  lines  which  correspond  to  the  non-radial  path 
of  the  stylus  or  inking  point. 

9.  Roulettes  or  Cycloidal  Curves.  Every  point  in  a 
rolling  curve  generates  a  periodic  curve  called  a  roulette. 
The  special  curve  which  results  depends  upon  the  shape 
of  the  generator  curve,  i.e.,  the  rolling  curve,  and  upon  the 
position  of  the  generatrix,  i.e.,  the  tracing  point  or  stylus, 
and  also  upon  the  shape  of  the  directrix,  i.e.,  the  guide  line 
or  track  of  the  generator  curve. 

When  the  directrix  is  a  straight  line  and  the  generator 
curve  is  a  circle  then  any  point  in  the  circumference  of  the 
latter  will  trace  a  cycloid  which  is  illustrated  by  AEGLN 
in  Fig.  150. 

The  equation  of  the  cycloid  is  given  in  (39)  and  (40) 


THE  USE  OF  POLAR  PAPER 


375 


in  which  r  is  the  radius  and  6  the  angle  of  rotation  of  the 
generator  circle. 


(39) 
(40) 

(41) 


x  =  rO  —  r  sin  8 
y  =  r  —  r  cos  8, 


x  =  r  cos 


ir-Jl_V2r^p, 


Fig.  150.— The  Cycloid. 

Eliminate  6  from  (39)  and  (40)  and  show  that  (41) 
and  (42)  result.  In  Fig.  150,  AN  is  laid  off  on  the  directrix 
XX  as  the  rectified  circumference  of  the  generator  circle 
whose  radius  is  r.  AN  is  subdivided  into  12  equal 
divisions  corresponding  to  the  subdivisions  of  the  circum- 
ference. The  center  of  the  generator  circle  will  be  directly 
above  each  of  these  twelve  numbered  positions  in  succession 
corresponding  to  each  twelfth  of  its  completed  rotation. 


376 


PRACTICAL  MATHEMATICS 


V  =  ( R+r )  cos  ft  -r  oos  6  (-&%£ ) 
»=(R+r)sin0-r  sin  Q  (-B-±T) 


Fig.  151.— The  Epicycloid. 


#=(R-r)cos0+rcos 
=(R-r)  sin 6—**  sin 


Fig.  152.— The  Hypocycloid. 


THE  USE  OF  POLAR  PAPER  377 

The  respective  positions  of  the  generatrix  are  A,  B,  C,  Z>, 
E,  F,  G,  H,  J,  K,  L,  M,  N.  The  generatrix  does  not  rise 
and  advance  uniformly  for  successively  equal  amounts  of 
rotation,  as  is  indicated  by  the  series  of  horizontal  parallels. 
10.  When  the  directrix  is  a  circular  arc  the  resulting 
curve  traced  by  the  point  A  may  be  either  an  epicycloid 
or  hypocycloid,  depending  upon  whether  the  generator 
circle  rolls  outside  or  inside  the  directrix.  The  epicycloid 
is  shown  in  Fig.  151,  and  its  equations  are  given  in  (43)  and 
(44).  The  hypocycloid  is  shown  in  Fig.  152,  and  its  equa- 
tions are  given  in  (45)  and  (46) .  R  and  r  are  the  respective 
radii  of  the  director  and  generator  circles.  Not  only  do 
the  generator  curves  rotate  but  they  also  revolve  about 
the  center  0.  The  amount  of  revolution  about  0  is  expressed 
by  6.  The  generator  circumference  is  divided  equally  into 
twelfths.  The  series  of  concentric  circles  is  drawn  through 
these  points  to  show  the  successive  elevations  of  the  gen- 
eratrix. The  arc  AN  of  12  ...  12  equals  the  circumfer- 
ence of  the  generator  circle  and  is  likewise  divided  into 
twelve  equal  divisions. 

(43)  y=  (R+r)  cos  0-r  cos  e(— 

(44)  x=(R+r)  sin  l-ran  8^ ~  . , 


(45) 
(46) 


y  =  (R  —  r)  cos  8+r  cos  0( j, 

x  =  (R  —  r)  sin  6  —  r  sin  0( j. 


When  the  generatrix  lies  outside  or  inside  of  the  periphery 
of  the  generator  circle  the  resulting  curve  will  be  looped 
and  is  known  accordingly  as  a  curtate  or  prolate  cycloid. 

Roulettes  are  applied  industrially  in  machine  design 
and  also  in  the  geometric  design  of  engraving. 


378 


PRACTICAL  MATHEMATICS 


Evolutes  and  Involutes.  Imagine  a  taut  string  un- 
wound from  a  bobbin  or  barrel  whose  cross  section  outline 
we  shall  call  an  evolute.  The  extremity  of  the  string  traces 
an  involute.  An  evolute  and  an  involute  are  two  related 
curves  in  so  far  as  one  depends  upon  the  other  in  its  construc- 
tion. The  tangents  to  the  evolute  are  normal,  i.e.,  perpen- 
dicular to  the  involute. 

Ex.  16.     Construct  the  involute  of  a  circle. 

11.  The  Trisection  of  an  Angle.  The  trisection  of  an 
angle  may  be  accomplished  by  means  of  the  conchoid  shown 


§^P^ 


'  /  i  i  i  \  \\\ 
'   I   i   \  x 


Fig.  153.— The  Conchoid. 


in  Figs.  153,  and  154.  XX'  is  an  axal  line  or  directrix  and 
0  a  pole  and  OP,  OJ,  OD,  etc.,  are  radii  vectors.  The 
conchoid  has  two  distinct  branches.  Each  segment  of  the 
radii  vectors  lying  between  the  directrix  and  the  curve 
equals  a  numeric  constant  called  the  modulus,  (a)  This 
property  enables  us  to  construct  the  curve  quickly,  (b) 
is  the  distance  of  the  pole  below  the  directrix.  The  equation 
of  the  conchoid  is  given  in  (49)  which  is  derived  from  the 
relation  of  the  sides  of  similar  right  triangles  123  and  022' 


THE  USE  OF  POLAR  PAPER  379 

as  expressed  in  (47)  and  (48).     The  rectangular  coordinates 
refer  to  the  axes  RS  and  XX'. 

(47)  ^JS 

y     }  QP     32' 

x       Va2  —  v2 


(48)  F+y  y       ■ 

(49)  •  x2y2(a2-y2)(y+b)2. 

It  is  customary  to  use  the  left  end  of  the  upper  branch 
of  the  conchoid  for  the  trisection  of  an  angle  as  shown  in 
Fig.  154.  The  angle  BOA  is  given.  Construct  PR ±0A 
and  fill  in  the  angular  space  with  radial  lines.  On  each 
radial  line  measured  outward  from  PR  lay  off  distances 
equal  to  20P.     Through  the  points  so  obtained  draw  the 

The  Conchoid. 


R  A 

Fig.  154.— The  Trisection  of  Angle  BOA 

conchoid  ADCEK.  Construct  PC  parallel  to  OA  intersect- 
ing the  conchoid  at  C.  Join  C  with  0  then  CO  A  =  \BOA . 
The  bisection  of  BOC  with  a  compass  completes  the  tri- 
section. 

It  is  not  necessary  to  reconstruct  the  conchoid  for  each 
angle  since  the  conchoid  may  be  laid  off  on  tracing  paper 
or  a  matrix  or  a  die  made  in  celluloid  or  metal.  If  the  vertex 
of  the  angle  0  and  one  of  its  sides  OB  are  brought  into 
coincidence  with  the  pole  0  and  axal  line  OC  respectively, 
as  shown  in  Fig.  155,  then  the  trisection  may  be  readily 
obtained  by  the  intersection  of  VP  a  parallel  to  OB  drawn 
from  V  the  intersection  of  the  directrix  with  OD  the  other 
side  of  the  angle. 


380 


PRACTICAL  MATHEMATICS 


12.  The  proof  of  the  trisection  of  an  angle  by  the  con- 
choid follows. 


Given  Z  AOB,  conchoid  CPP'D  with  pole  0  and  directrix 
XX 

Prove  ZPOC^f?. 

O 

(1)  Construct  7P  HOC;  1 

NM  ||  VX  through  M,  midpoint  UP;    \ Cons.  Ax. 

join  V  with  M  forming  Z2'.  J 

(2)  UVP  is  a  Rt.  A (1)  Def.  Rt.  A 

(3)  .*.    UM  =  MP  =  VM  =  OV Rt.  Z  insc.  O 

(4)  .*.   As  VMP,  and  OVM  are  isosceles-Def.  Isos.  A 

Cons.  Theo.  Isos.  A 


(5)  /.    Zsl  =  l,,  =  land2  =  2/- 

(6)  but  Z2,  =  1,  +  1,,  =  2Z1  — 

(7)  .*.      Z2  =  2Z1 


(8)  /.      /.P0C  = 


A  AOB 


Alt.  int.  Z  s. 
Ext.  Z  A,  subs. 

-(5)(6)=ty.Ax. 

-Subs.  Ax. 


THE  USE  OF  POLAR  PAPER 


381 


Ex.  17.     By  means  of  a  conchoid  trisect  a  radian  of  angle. 

13.  The  Multisection  of  an  Angle.  An  angle  may  be 
multisected,  i.e.,  divided  into  any  number  of  divisions  by 
means  of  the  chordel  which  is  illustrated  in  Figs.  155  and 
156.  In  Fig.  155,  0  is  the  pole  and  OB  the  directrix  or 
axal  line.  The  chordel  is  so  named  from  the  fact  that  the 
intercepted  arcs  are  evenly  divided  by  applying  consecu- 
tively a  chord  or  element  of  definite  length.  In  Fig.  155, 
the  element  is  used  seven  times  and  serves  to  divide  an 


R-2csc2„ 

n  =  number  of  \  elements 

.— - --_       ^  *  chords   n 


\    \ 


H»\- 


9Y2\5  \  \  \  \  \  \| 

5&^"V4s\  \  \  \  \  \% 


154-8  41  II  4-*" 

=  lengthen  chord  or  element 


j i—  leugiu  oi  unoru  or  ei€ 

Jl  j  Id  g|e      !     B-f-t 


hi 


y  /ill  TiH  i    i    i  i    I 

■   y  /  /  / 
/  / 


i  i  i  i         1 
/  /  /  i  i  l  i       f 


///  //  /ff/ff 
'  y  /  /  /  /  /  /  /  / 


Fig.  155.— The  Chordel. 


angle  into  seven  equal  parts.  In  Fig.  156,  the  element  is 
used  three  times  and  serves  to  divide  an  angle  into  three 
equal  parts. 

The  chordel  is  constructed  by  beginning  at  the  directrix 
and  laying  off  consecutive  chords  on  the  concentric  arcs  so 
that  there  will  be  as  many  elements  as  the  required  numeric 
division  of  the  angle.  A  smooth  curve  is  then  passed 
through  the  points.  The  chordel  is  symmetrical  to  the 
directrix  and  has  one  less  convolution  than  the  number  of 


382  PRACTICAL  MATHEMATICS 

elements.  The  equation  of  the  chordel  is  given  in  (50)  in 
which  I  is  the  length  of  an  element,  n  the  number  of  elements 
and  R  and  0  are  polar  coordinates* 

(.50)  R=i™i- 

In  Fig.  156,  the  angle  BOA  is  to  be  trisected  by  the 
chordal.  Regard  the  vertex  0  as  the  pole  of  the  chordal 
and  the  side  OA  as  the  directrix.  Construct  three  con- 
centric arcs  and  lay  off  a  convenient  element  three  times. 


O— = p  A 

Fig.  156.— The  Trisection  of  BOA,  by  Means  of  the  Chordel. 

Through  the  last  points  of  division  draw  the  chordal  CED 
intersecting  the  angles'  side  OB  at  E.  Draw  another  con- 
centric arc  EF  through  E,  and  observe  that  the  element 
divides  the  arc  EF  evenly  at  G  and  H.  Connect  G  and  H 
with  0.     Then 

ZsE0G  =  G0H  =  H0F  =  ^-. 


Ex.  18.     By  means  of  the  chordel  divide  an  angle  of  100°  into 
7  parts. 

Ex.  19.    By  means  of  the  chordel  divide  a  radian  into  5  parts. 


CHAPTER  XX 
SOLVING   FORMULAS   BY   CHARTS 

1.  The  graphic  solution  of  many  formulas  may  be 
obtained  by  preparing  a  suitable  chart  or  diagram  of  lines 
which  may  be  constructed  on  plain  paper.  A  straightedge  or 
stretched  thread  is  applied  upon  the  chart  and  its  inter- 
section with  a  fixed  graduated  line  called  an  index  or  sup- 
port gives  the  desired  solution. 

2.  The  chart  solution  of  a  quadratic  equation  is  shown 
in  Fig.  157,  in  which  the  index  is  a  parabolic  curve.  The 
equation  of  the  second  degree  for  the  variable  to  is  expressed 
in  (1). 

(1)  a)2+ww-r-tf  =  0. 

The  horizontal  line  through  the  center  of  the  chart  is 
called  the  zero  line.  The  left  and  right  vertical  bounding 
lines  are  designated  as  the  u  and  v  axes  respectively.  The 
axes  are  uniformly  graduated  for  both  positive  and  negative 
values.  To  solve  a  quadratic  equation  observe  the  u  and 
v  values  of  the  given  equation.  Locate  these  values  on 
their  respective  axes  and  join  them  with  a  straightedge. 
The  intersection  of  the  straightedge  with  the  index  indicates 
the  solution  for  w. 


Ex.  1.    Verify  the  following  solutions: 

Given 

Solution 

(2)          o)24-.5o>-l.o=0, 

(0  =  1, 

(3)         w2 -.75(0-2.5=0, 

a>=2, 

(4)          to2 -2.5(0  +  1.5=0, 

(o  =  15. 

383 


384 


PRACTICAL  MATHEMATICS 


2\ 


1.5 


1U 


-1 


1.5 


■2.5 


\\ 


u"+wtt+v=0 
Read  M  and  V  on  the  respective 
axes  and  join  the  points  by  a 
line  or  thread.  The  value  of 
10  is  read  off  from  the  curve 
where  the  thread  intersects 
the  curve. 


1.5 


1.5 


1.5 


r«J 


Fig.  157. — Chart  Solution  of  a  Quadratic  Equation. 


SOLVING  FORMULAS  BY  CHARTS  385 

3.  The  construction  of  the  index  is  obtained  by  locating 
each  of  its  successive  points  of  graduation  by  two  intersect- 
ing lines.  Suppose  we  wish  to  locate  the  point  w  =  l.  Sub- 
stitute 1  for  to  in  (1)  which  reduces  to  (5). 

(5)  l+u+v  =  0. 

In  (5)  u  and  v  may  have  any  pair  of  associated  values 
which  will  validate  it.  Therefore,  when  u  =  0,  then  v  =  —  1, 
and  when  v  =  0,  then  u  =  —  1.  Locate  these  pairs  of  points 
on  the  u  and  v  axes  respectively,  and  join  them  by  straight 
lines.     Their  intersection  gives  the  point  co  =  1. 

In  order  to  locate  any  other  point  on  the  index  such  as 
w  =  2  substitute  2  for  w  in  (1)  which  reduces  to  (6). 

(6)  4+2^+^  =  0. 

In  (6)  when  v  =  0,  then  u=—  2,  and  when  u  =  0,  then 
v  =  —  4.  Draw  straight  lines  through  these  pairs  of  points. 
Their  intersection  locates  the  point  o>  =  2. 

Other  points  on  the  index  are  determined  in  the  same 
manner.  By  extending  the  index  to  the  right  of  the  v  axis 
negative  value  of  w  may  be  determined. 

4.  If  the  constants  u  and  v  are  too  large  to  be  read  on 
the  chart  then  divide  u  by  10  and  v  by  100  and  multiply 
the  index  reading  by  10.  Thus  to  solve  (7)  rewrite  it  as 
(8)  and  (9).  Solving  (9)  by  the  chart,  gives  g>  =  .6375  and 
therefore  A  =  10 w  =  6.375. 

(7)  A2 +25 A-  200  =  0 given  equation. 

(8)  100g>2+250w-  200  =  0 subs.  A  =  10w. 

(9)  /.      w2+2.5co-  2  =  0 div.  (8)  by  100. 

5.  The  chart  solution  for  a  cubic  equation  is  shown  in 
Fig.  158,  in  which  the  index  is  also  a  parabolic  curve.  Every 
equation  of  the  third  degree  may  be  reduced  to  the  form 

(10)  in  which  the  second  power  of  the  unknown  is  missing. 


386 


PRACTICAL  MATHEMATICS 


w3-f-t?w+u=0 

It  Locate  the  values  of  u  and  v  on  the  respective  axes  V 

join  these  points  with  a  thread  intersecting  the  curve  at  value  M 


Fig.  158.— Chart  Solution  of  a  Cubic  Equation. 


SOLVING  FORMULAS  BY  CHARTS  387 

Locate  the  constants  u  and  v  of  the  given  equation  on 
their  respective  axes  in  Fig.  158,  and  join  them  by  a  straight 
line.  The  intersection  of  the  latter  with  the  index  indicates 
the  solution  of  (10).  Only  negative  values  of  u  and  v  are 
shown  in  the  chart. 

(10)  w3+*;w+w  =  0. 
Ex.  2.     Verify  the  following  solutions: 

Given  Solution 

(11)  (o3-    3(o-    2=0,  (o=2, 

(12)  w3+  .5(0-1.5=0,  w  =  l, 

(13)  (o3-. 75o> +.25=0,  (o  =  .5. 

6.  If  the  numeric  values  of  u  and  v  are  too  large  to  be 
read  on  the  chart  then  divide  v  by  100  and  u  by  1000  and 
multiply  the  index  reading  by  10. 

Thus  to  solve  (14)  rewrite  it  as  (13)  and  since  w  =  .5, 
then  ?/ =  10o)  =  5. 

(14)  y3  —  Iby + 250  =  0 — given  equation. 

(15)  (10g))3-75(10w) +250  =  0 subs.  y  =  10w  in  (14). 

(16)  /.   lOOOto3- 750w+250  =  0 simplifying  (15). 

(13)         /.      w3-.75(o+.25  =  0 div.  (16)  by  1000. 

7.  In  order  to  reduce  a  complete  cubic  equation  (17) 
to  the  required  form  (10)  substitute  w  —  —    for  z  where  k 

o 

is  the  coefficient  of  z2.     Reduce  the  resulting  equation  (18) 
to  (19)  which  is  identical  with  (10). 

(17)  z3+fc«2+fe+w  =  0, 

(18)  ^-iV+^('a)-^2+^(o-|-)+m  =  0, 


3/     'V        3 


(19) 


(10)       (o3+^(o+w  =  0. 


388 


PEACTICAL  MATHEMATICS 


8.  The  charts  represented  by  Figs.  157  and  158,  suggest 
a  method  of  plotting  three  variables  in  the  plane  of  the  paper. 

The  charting  of  reciprocal  formulas  which  are  represented 
by  (20)  and  (21)  is  shown  in  II  of  Fig.  159. 

(20) 


(21) 


±=L+L 
R     rtra! 

C    C\     C2 


Equation  (20)  is  the  formula  for  joint  resistance  R  of 
two  resistances  ri  and  r2  connected  in  parallel. 

Equation  (21)  is  the  formula  for  equivalent  capacity  C 
of  two  condensers  which  are  placed  in  series.  The  individual 
capacities  are  C\  and  C2. 


Fig.  159. — The  Charting  of  Three  or  More  Variables. 

9.  In  Fig.  159,  chart  II  represents  two  oblique  axes  on 
which  ri  and  r2  are  read.  The  angle  formed  by  n  and  r2 
is  bisected  by  the  index  which  gives  values  of  R.  The 
three  scales  are  uniform  and  equally  spaced.  II  may  be 
used  for  (21)  or  any  similar  reciprocal  formula  by  substitut- 
ing the  corresponding  elements  for  ri,  r2  and  R.  It  is  most 
convenient  to  construct  the  axes  at  an  angle  of  120°.  In 
the  figure  the  straight  line  AB  joining  the  point  A  for  which 
n  =  20  with  the  point  B  for  which  r2  =  15  intersects  the 
index  at  the  point  C  for  which  #  =  8.57  when  these  values 
are  substituted  in  (20)  we  obtain  (22). 

117        1 
(22)  20+l5  =  60  =  8J37' 


SOLVING  FORMULAS  BY  CHARTS  389 

The  proof  of  the  chart  solution  follows:  join  C  with  the 
10  mark  on  the  n  and  r2  axes  and  designate  these  respective 
points  by  D  and  E.  Then  triangles  ADC  and  CEB  are  simi- 
lar and  triangles  DCO  and  COE  are  equal  equilateral  tri- 
angles.    Therefore, 

(23)  CO  =  DC  =  CE  =  OD  =  OE. 

feiA.     AD     CE    AD  .        . ■'  ' 

(24)  -R7^=T777  =  ^F^ homol.  sds.  sim  As. 

DC     EB     DO 

and 

,oc,     AD+DO    CE+EB  

(25)  — ^ttt —  =  — wb accretion  proportion. 

uu  hits 

(26)  AD+DO  =  n,        CE+EB  =  OE+EB  =  r2, 

and 

EB  =  r2-R Mag  Ax. 

(27)  5=^ Subs'  in  (25)* 

(28)  rir2  =  R(ri-\-r2) prod,  means,  trans. 

(29)  4  =  -+^ divAx. 

R     r      r2 

Therefore  any  equation  of  the  form  (28)  or  (29)  or  its 
equivalent  may  be  charted  as  shown  in  II,  Fig.  159.  How 
can  II  be  supplemented  to  comprehend  an  indefinite  number 
of  elements. 

10.  The  charting  of  linear  formulas  which  are  represented 
by  (30)  or  (31)  is  shown  in  I,  Fig.  159. 

The  variables  u  and  v  are  scaled  on  the  like  named 

vertical  axes  which  are  separated  by  the  distance  c.     The 

variable  s  is  scaled  on  the  index  which  is  separated  from 

the  u  and   v   axes    by   distances   which  are  in  the  ratio  of 

a:b.    For  (31)  t  may  be  read  on  the  index,  but  the  values  of 

d 
s  are  -  greater  than  those  for  t. 
c 

(30)  av-\-bu  =  cs, 

(31)  av+bu  =  cs+d  =  tc. 


390  PRACTICAL  MATHEMATICS 

Construct  a  parallel  to  c  passing  through  C,  then  by 
similar  triangles, 

(33)  .'.     av-\-bu=(a+b)s  =  cs. 

11.  Equations  (34)  and  (35)  reduce  to  the  linear  forms 
(36)  and  (37),  which  are  comparable  with  (30)  and  (31). 

(34)  W  =  PR, 

05)  ijg,; 

(36)  2\ogI+\ogR  =  \ogW, 

(37)  3  log  d+log  b  =  \og  J-f  log  12. 

In  preparing  the  chart  for  (36)  the  u  =  log  d  and  z;  =  log  b 
axes  and  the  index  =  log  /  are  graduated  logarithmically  as 
shown  in  I,  Fig.  159.  The  index  is  located  by  dividing 
the  distance  c  into  two  segments  whose  ratio  2:1  cor- 
responds to  the  ratio  of  the  coefficients  of  log  d  and  log  b 
in  (36). 

12.  If  the  ranges  of  values  of  the  variables  are  different 
then  different  scales  may  be  used  according  to  the  relation 

(38)  where  h,  h,  and  Z3  are  the  respective  scales  for  log  /, 
log  R,  and  log  W  or  their  equivalents.  In  charting  (36), 
the  initial  marks  on  the  two  axes  and  index  read  1 
to  correspond  to  log  1=0.  In  (37)  the  initial  marks  for  the 
b  and  d  scales  read  1  but  for  the  /  scale  the  initial  mark 
reads  .0833.  Therefore  the  1  mark  on  the  index  will  be 
above  the  corresponding  1  mark  on  the  axes. 

(38)  '        hh 


fort 
sion  and  sag  in  wire  spans. 


Ex.  3.     Prepare  charts  for  the  following:  D  =  . 0015ml  — .   Ten- 


SOLVING  FORMULAS  BY  CHARTS  391 

Ex.  4.     log  e  =a  log  t  +  $.     Thermo-electric  couple. 

Ex.5.     /=— .     Ohm's  law. 
H 

Ex.6.     Rt=R0(l+aT).   Temperature  coefficient. 

e  (n 

Ex.  7.     I  =7r\h^~-      To  find    the    greatest  current  I  from    a 
I  \  Hz 

given  number  of  cells  through  an  external  resistance  R. 
Ex.  8.     P=EI  cos  <J>.     Power  in  A.C.  circuit. 

Ex.  9.     X  =rI  —  —  1j.     Measurement  of  high  resistance. 

ft 

Ex.  10.     B  =  1317a  /—  +H.  Traction  method  of  magnetic  test. 


Ex.  11.    Z  =  VR2  +  W2L2.     Impedance  formula. 

In  the  preceding  charts  it  has  been  shown  that  the  index 
may  be  straight  or  curved  and  the  axes  may  be  parallel  or 
oblique.  Another  case  arises  in  which  the  index  joins  the 
extremities  of  the  axes  forming  a  Z  chart  as  shown  in  III, 
Fig.  159.  The  scales  of  axes  A B  and  DG  are  reversed  so 
that  the  respective  zeros  are  at  A  and  D.  From  the  relation 
of  the  parts  any  of  the  following  equations  or  equivalents 
may  be  solved  by  the  Z  chart.     K  =  constant  =  c-\-d. 


(39) 
(40) 


When  the  number  of  variables  exceeds  three  it  is  possible 
to  decompose  the  given  equation  so  that  the  composite 
chart,  i.e.,  chart  of  charts  may  be  constructed  wrhich  provides 
for  a  solution  through  successive  operations.  In  such  cases 
there  would  be  a  number  of  indexes  which  wTould  in  turn 
serve  as  an  axis  to  establish  a  reading  on  the  next  index. 


a 
V 

_b 
c 

e 

c+d 
~a+b~~ 

e—c 

c- 

-d 

-a 

a+b  = 

•;-*■ 

392  PRACTICAL  MATHEMATICS 

Ex.  7  may  be  solved  by  a  composite  chart  constructed 
as  follows  if  all  the  elements  are  variable.  Decompose  the 
equation  as  follows: 

IW       In 

(41)  I-E^-E^m. 

Construct  seven  vertical  lines  graduated  logarithmically 
and  designated  R,  r,  I,  E,  N,  i*i,  12.  The  scales  of  R,  r,  N,  i\, 
will  be  twice  those  of  I,  E  and  t*2.  Dispose  R,  r  and  i\  so 
that  i\  indicates  the  product  Rr.  Dispose  12  and  N  with 
regard  to  i\  so  that  i%  indicates  the  ratio  of  N  to  i\.  Dispose 
/  and  E  with  regard  to  1*2  so  that  /  indicates  the  product  Ei2> 


CHAPTER  XXI 
MEASUREMENT    OF   ANGLES 

1.  Angles  are  measured  usually  with  a  protractor  which 
is  graduated  in  degrees  of  arc.  We  read  degrees  of  arc  in 
measuring  an  angle  but  express  the  answer  in  degrees  of 
angle  because  an  angle  is  measured  by  its  intercepted  arc. 

A  degree  is  one-ninetieth  of  a  right  angle.  The  four 
quadrants  contain  each  90  degrees  of  angle  and  90  degrees 
of  arc. 

A  radius  vector  in  making  a  complete  rotation  describes 
a  perigon  or  360  degrees  of  angle,  as  well  as  360  degrees  of 
arc. 

Ex.  1.     How  many  perigons  are  described  in  1260°,  1000°,  810°? 

The  ratio  of  the  circumference  to  the  radius  of  a  circle 

is  a  constant  of  nature  and  is  abbreviated  by  2x. 

22 
2x  =  2Xy   approx.  =  2X3.14159   approx.  =  6.28   approx. 

44  25 

Ex.  2.     What  is  the  error  in  choosing  — ,  also  —  for  the  value 

of  2x? 

2.  If  we  were  to  circularize  the  radius  of  a  circle,  i.e., 
bend  it  to  the  form  of  the  circumference  then  the  latter  would 
contain  the  radius  2tc  times. 

The  portion  of  the  circumference  exactly  equaling  the 
length  of  the  radius  is  called  a  radian  or  a  radian  of  arc. 
The  central  angle  subtended  by  a  radian  of  arc  is  also  called 
a  radian,  i.e.,  a  radian  of  angle. 

393 


394  PRACTICAL  MATHEMATICS 

360°  corresponds  to  2x  radians  =  6.28  radians. 

360°     360°     c„00  .       ■■. 

One  radian  =  — —  =  ^-7^  =  57.3    approximately. 
zx       0.Z0 

The  number  of  radians  may  be  expressed  in  integers 
or  in  multiples  of  x.  The  former  is  designated  the  unit 
measure  and  the  latter  the  x  measure  of  the  angle. 

In  any  expression  for  the  measure  of  an  angle,  x,  3.14  or 
180°  are  equivalent  and  therefore  may  replace  one  another. 
When  an  angle  is  expressed  in  unit  measure  multiply  the 
units  by  57.3°  and  when  expressed  in  x  measure  replace 
x  by  180°,  in  order  to  obtain  its  degree  measure.     When 

the  angle  is  expressed  in  degrees  multiply  it  by  — —  or  — '— 

180        180 

to  express  it  in  x  or  unit  measure  respectively. 

Ex.  3.  Determine  the  radian  measure  of  the  following  angles 
both  for  unit  and  %  equivalents:  1°,  2°,  3°,  4°,  5°,  10°,  15°,  30°, 
45°,  60°,  75°,  90°,  105°,  120°,  135°,  150°,  165°,  180°,  270°,  360°, 
540°.     Tabulate  the  three  equivalents  for  each  angle. 

Ex.  4.  Determine  the  degree  equivalent  for  — ,  — ,  — ,  — -,  — , 
c  5     5     4     4     4 

~,   O.lx,  1.5x,  0.75x. 

o 

Ex.  5.  Determine  the  degree  equivalent  for  the  following 
angles:   1.5,  2,  1.5708,  4.7124. 

Ex.  6.  Express  in  unit,  x  and  degree  measure  the  sums  indi- 
cated with  mixed  notation  below: 


(a) 

it +6, 

(h) 

30° +107°, 

(b) 

3+2' 

w 

30°+^, 

(c) 

1+2.5, 

U) 

180° -0.6, 

(d) 

30° +2,1, 

(*) 

90°+0.2, 

(«) 

45       3' 

(I) 

90°+tan  _1  (1), 

(/) 

6.2-15°, 

(m) 

sin  -1(0.5)+60°, 

to) 

0.7x+0.7, 

(n) 

2x  +57.3°  +2.718. 

MEASUREMENT  OF  ANGLES  395 

Example  (a)  may  be  written, 

x+6  =3.14+6  =9.14  (radians), 
x  +6  =  1S0°  +6X57.3°  =523.8°, 

523.8° X— X— =2.91x  (radians). 

lot) 

The  unit  equivalent  of  angles  from  1°  to  90°  is  given  in  Table  VIII, 
page  121,  under  the  column  headed  radians. 

3.  Trigonometric  formulas  are  verified  by  geometric 
proof  and  also  by  establishing  identities  in  the  equations 
through  the  substitution  of  recognized  elementary  trig- 
onometric relations. 

Thus  (1),  (2),  (3)  may  be  established  by  constructing 
any  angle  A.  Its  radius  vector,  perpendicular  and  pro- 
jection are  designated  by  RV,  1  and  proj  respectively. 
Then, 

(a)  _L2+proj2  =  RV2 law  of  Rt  A. 

i   2  *2 

(6)  i^+ E^r  - 1 div.  (a)  by  RV2. 

V  RV2    RV2  y 

I  2  proi2 

(c)     but    — r-5  =  sin2  A  and 0  =  cos2  A — def .  sin  and  cos. 

RV2  RV2 

(1)  .*.        sin2  A+cos2  A  =  1 sub.  (c)  in  (b). 

(2)  /.       tan2  A  + 1  =  sec2  A div.  (1)  by  cos2  A. 

(3)  /.       l+-cot2  A  =  csc2  A div.  (1)  by  sin2  A. 

Other  elementary  forms  are  given  in  (4),  (5)  and  (6); 
vers  is  the  abbreviation  for  versine  and  covers  is  the  abbre- 
viation for  coversine. 

/a\  sinA     ,        . 

(4)  j=  tan  A. 

v  cos  A 

(5)  vers  A  =  1  —  cos  A, 

(6)  covers  A  =  1  —  sin  A . 


396  PRACTICAL  MATHEMATICS 

Ex.  7.     Solve  (1)  .  .  .  (6)  for  each  function  which  appears  in 
it. 

Ex.  8.     Verifying  the  following  equation : 

.  ,  tan  0  +sec  0  —  1 

{a) =tan  6+sec  6. 

tan  8— sec  6  +  1 

(6)     .*.     tan  8+sec  6-1  =tan2  6-sec?  6+tan  6+sec  6 


clearing  frac.  in  (a). 

(c)  .*.      -1  =tan2  6  -sec2  6 simplifying  (b) 

(d)  :.     tan2  6+1  =sec?  6 trans,  in  (c) 

Equation  (a)  is  verified  by  reducing  it  to  the  elementary  form  (2). 

Ex.  9.  By  means  of  (1)  .  .  .  (6)  either  singly  or  in  combina- 
tion every  trigonometric  function  may  be  expressed  in  terms  of 
every  other  trigonometric  function.  Express  all  functions  of  A  in 
terms  of  sin  A. 

4.  Functions  of  Algebraic  Sums  of  Angles.  When 
two  angles  A\  and  A 2  are  united  by  a  plus  or  minus  sign 
then  the  sin  and  cos  of  such  a  combination  is  expressible 
in  terms  of  the  sin  and  cos  of  the  constituent  angles  as  shown 
in  (7),  (8),  (9)  and  (10).  The  left-  and  right-hand  members 
are  called  the  contraction  and  expansion  respectively. 

(7)  sin  (A1+A2)  =sin  A\  cos  A2+COS  A\  sin  A2, 

(8)  sin  (Ai~ A2)  =sin  A\  cos  A2— cos  A\  sin  A2, 

(9)  cos  (A1+A.2)  =cos  A\  cos  A2— sin  A\  sin  A2, 

(10)  cos  (Ai— A2)  =cos  A\  cos  A2+SU1  A\  sin  A2. 
Ex.  10.    Write  the  expansion  for  the  following: 

(a)  sin(6+<J)),  (c)  sin  (6x+c), 

(b)  cos(cl>-a),  (d)  cos  (ut+K). 
Ex.  11.    Write  the  contraction  for  the  following: 

(a)  sin  6  cos  4>  -cos  8  sin  <}>. 

(b)  cos  8  cos  4>  —sin  6  sin  <{>. 


MEASUREMENT  OF  ANGLES  397 

5.  Functions  of  Multiple  Angles.  When  Ai  =  A2  =  A 
then  (7)  and  (9)  reduce  to  (11)  and  (12)  respectively.  The 
latter  express  the  relation  between  an  angle  and  its  second 
multiple. 

(11)  sin  2A=  2  sin  A  cos  A, 

(12)  cos  2A  =  cos2  A  —  sin2  A . 

(11)  serves  to  indicate  that  the  multiplier  of  an  angle 
cannot  be  factored  and  written  as  a  multiplier  of  the  function, 
i.e.,  sin  2A^2  sin  A. 

Ex.  12.    Verify  (11)  by  substituting  A  =60°,  45°,  120°,  x. 

6.  Functions  of  Half  Angles.  By  adding  (1)  and  (12) 
we  obtain  (13),  by  subtracting  (12)  from  (1)  we  obtain  (14). 


(13)  cosA=J 


1+ cos  2 A 

o  » 


/n\  •     a        /I-  cos  2A 

(14)  sinA  =  ^ g • 

A 

Ex.  13.     Substitute  —  for  A  in  (13)  and  (14)  and  simplify. 

7.  Tangent  Functions.  The  tangent  functions  are 
derived  from  the  corresponding  sin  and  cos  functions  by 
division  and  simplification. 

.._  /a    *  a\      tanAi'±tanA2 

(15)  tan  (Ai±A2)  =  T-n i~i 7"> 

v     J  v  ■         l±tanAitanA2 

/i*\  4-      c>a      2  tan  A 

(16)  tan  2 A 


(17)  tan  A  = 


l+tan2A' 
1  —  cos  2A  _  sec  2A  —  1 
sin  2A  tan  2A 


Ex.  14.  Derive  (15)  from  (7)  and  (9)  for  tan  (Ax+Ai)  and 
from  (8)  and  (10)  for  tan  (A^A*). 

Ex.  15.     Derive  (16)  from  (11)  and  (12). 

Ex.  16.  Derive  (17)  from  (13)  and  (14),  also  derive  (16)  from 
(15). 

A 

Ex.  17.     Substitute  -  for  A  in  (16)  and  (17)  and  simplify. 


398  PEACTICAL  MATHEMATICS 

8.  Conversion  Formulas.  Another  group  of  useful 
formulas  is  given  in  (18),  (19),  (20)  and  (21).  This  group 
provides  a  means  for  changing  an  algebraic  sum  into 
a  product. 

(18)  sin^.i+sin^.2  =  2  sinJ(Ai+A2)  cos  K^-i-  A2), 

(19)  sin  A i  —  sin  A2  =  2  cos  \(A\+A2)  sin  \{A\—  A2), 

(20)  cos  Ai+cos^4.2  =  2  cos  %(Ai+A2)  cos  J(Ai  — A2), 

(21)  cos  A\  —  cos  A2  =  — 2sini(Ai+A2)  sin  i(Ai  —  A2). 

Ex.  18.  Multiply  (18)  by  (19)  and  determine  the  simplified 
value  of  sin2  Ax  —  sin2  A2. 

Ex.  19.  Multiply  (20)  by  (21)  and  determine  the  simplified 
value  of  cos2  Ai  -cos2  A*. 

9.  Exponential  and  Trigonometric  Series.     (22)  and  (23) 

express  the  values  of  the  exponentials  expanded  into  an 
infinite  series.      In  (22)  every  term  is  positive,  whereas  in 

(23)  the  terms  are  alternately  positive  and  negative. 

!  is  the  factorial  symbol  and  means  the  product  of  all 
numbers  from  1  to  and  including  the  indicated  number. 

«2       2;3  4  5  q 

(22)  ^=1+2+2!+3!+i!+5!+6!+--- 

22        2«3        24        zh        2^ 

(23)  ,-.,l-«+5__+£-_+g_.  .  . 

(23)  is  the  reciprocal  of  (22)  and  may  be  obtained  by 
division,     sin  z  and  cos  z  are  also  expanded  into  the  series 

(24)  and  (25)  respectively. 

(24)  ,  sinz  =  z-^+!j-^+.  .  . 

z2     z4     ZG 

(25)  cosz=1~2!+4!"6!  +  -  '  ' 


MEASUREMENT  OF  ANGLES  399 

It  will  be  observed  that  (24)  and  (25)  may  be  built  from 
(22)  and  (23)  by  the  selection  of  definite  terms. 

One-half  the  difference  between  (22)  and  (23)  is  written 
as  (26)  which  is  called  the  hyperbolic  sine  and  is  abbreviated 
sinh  z.  One-half  the  sum  of  (22)  and  (23)  is  written  as  (27) 
which  is  called  the  hyperbolic  cosine  and  is  abbreviated 
cosh  z.  The  ratio  of  (26)  to  (27)  is  written  (28)  which  is 
called  the  hyperbolic  tangent  and  is  abbreviated  tanh  z. 
The  reciprocals  of  sinh,  cosh  and  tanh  are  csch,  sech  and 
coth  respectively. 

£2_£-z  ^       Z5       Z7 

(26)  sinh  2=-j-  =*+3!+5i+7!+-  •  ■ 

(27)  cosh*  =  ^  =  l+J+g+g+... 
,__:  ,         sinh  z     sz—e~z 

(28)  tanh^35ST2=^+^- 

Ex.  20.  The  following  formulas  are  used  for  the  determination 
of  the  electromotive  force  and  current  required  at  the  transmission 
end  of  a  track  circuit  used  for  signaling.  Simplify  the  formulas 
by  substituting  hyperbolic  functions  to  replace  the  exponential 
notation. 

(b)     »-A/Wn-e-*>fcW 

2Vjt1\  / 

Calculate  e  and  i  for  the  following  values  rx  =.9,  r2  =2.5,  Ex**\, 
ii  =  l  when  05  =  1,  2,  3,  4,  5  .  .  .  15. 


400  PRACTICAL  MATHEMATICS 

10.  The  hyperbolic  functions  are  related  by  the  follow- 
ing formulas. 

(29)  cosh2  z-sinh2  2  =  1, 

(30)  l-tanh2z  =  sech2z, 

(31)  coth2z-l  =  csch2z, 

(32)  sinh  (z±u)  =  sinh  z  cosh  u  =bcosh  z  sinh  u, 

(33)  cosh  (z± u)  =  cosh  z  cosh  wisinh  z  sinh  u,  ■ 

Ja  i    /    ,     n      tanh  z=btanh  w 

(34)  tanh(^«)-1±tanh.tenhw. 

The  numeric  values  of  hyperbolic  functions  are  obtained 
from  Table  XXIII,  page  366. 

Ex.  21.  Derive  the  formulas  for  sinh  2z,   cosh  2z,  tanh   2z, 

2  Z 

sinh  -,  cosh  — . 

Ex.  22.  In  (22)  substitute  2  =  1  and  calculate  e  by  adding  the 
numeric  values  of  the  first  twelve  terms  carried  to  8  decimal 
places. 


CHAPTER  XXII 


HARMONIC   MOTION 


1.  A  point  which  moves  uniformly  in  a  circle  projects 
orthogonally  into  a  point  which  moves  non-uniformly 
along  any  diameter  of  the  circle.  The  latter  motion  is 
called  simple  harmonic  motion  and  is  definite  because  it 
obeys  a  precise  law. 

In  Fig.  160  the  arm  CP  rotates  about  the  center  0. 
When  the  arm  rotates  with 
uniform  angular  velocity,  then 
every  point,  such  as  P,  in  the 
arm  CP  rotates  with  like  uni- 
form angular  velocity.  There- 
fore, N  and  M  the  respective 
orthogonal  projections  of  P 
on  the  vertical  and  horizontal 
diameters  move  with  simple 
harmonic  motion.  The  dis- 
placement of  N  from  the  center 
C  is  CN  and  is  numerically 
equal  to  the  length  of  the  arm 
CP  multiplied  by  the  sine  of 

6,  the  angle  of  its  rotation  measured  from  the  initial  posi 
tion  CO,  as  expressed  in  (1). 


/     TN 

.._\p 

/             Y 

7                   \ 

(    ,c 

A        :         1 

M 

Fig.  160. — The  Projection  of 
Uniform  Circular  Motion  on 
a  Diameter. 


(1) 


CN  =  CP  sine. 


(1)  may  be  abbreviated  as  (2)  by  designating  the  vertical 
displacement  by  y  and  the  arm  by  R. 


(2) 


=  12  sine. 


401 


402  PEACTICAL  MATHEMATICS 

The  greatest  value  of  y  equals  R  and  corresponds  to 
the  displacement  of  N  when  the  arm  CP  is  in  a  vertical 
position.  Then  P  coincides  with  N  and  the  angle  0  =  90°. 
These  facts  may  be  verified  by  substituting  0  =  90°  in  (2) 
which  becomes  y  =  R  sin  90°  =  R. 

When  the  arm  CP  rotates  beyond  90°  the  displacement 
decreases  so  that  when  0  =  180°  the  value  of  y  is  zero.  The 
rotation  of  CP  beyond  180°  gives  a  displacement  of  N  below 
C  and  the  corresponding  value  of  y  is  regarded  with  a 
negative  sense.  The  greatest  negative  value  of  y  equals 
— R  and  occurs  when  the  arm  CP  has  rotated  through  270°. 
When  the  arm  CP  is  rotated  from  270°  to  360°  the  value 
of  y  decreases  and  is  numerically  equal  to  zero  when  0  =  360°. 
If  the  rotation  of  CP  were  continued  beyond  360°,  then 
for  each  additional  revolution  of  P  the  simple  harmonic 
motion  would  be  repeated  in  the  same  manner  and  in  the 
same  order  as  described  above. 

2.  The  Sine  Curve  or  Record  of  Simple  Harmonic  Motion. 
A  continuous  record  of  the  simple  harmonic  motion  of  the 
projection  of  P  is  shown  in  the  sine  curve  of  Fig.  161. 
The  whole  arc  of  the  circle  OBFD  is  rectified  by  extending 
the  horizontal  diameter  from  0  to  E.  Then  OE  corresponds 
to  the  circumference  or  360°  of  arc.  The  ordinates  of 
the  sine  curve  represent  the  corresponding  displacements  of 
the  projection  of  P  at  its  successive  positions  1,  2,  3,  .  .  22, 
23,  24.  One  complete  rotation  of  CP,  i.e.,  one  complete 
revolution  of  P,  produces  one  cycle  of  change  extending 
from  0  to  E  or  from  0°  to  360°  or  from  0  to  2%  radians. 
In  this  way  simple  harmonic  motion  is  unfolded  into  a 
rectilinear  representation.  This  process  is  comparable  to 
changing  the  rotation  of  a  crank  arm  of  an  eccentric  of  an 
engine  into  the  rectilinear  motion  of  the  slide  of  a  valve. 

Ex.  1.     Construct  the  sine  curve  by  projection  as  follows: 
Construct  a  unit  circle  with  center  C.    A  unit  circle  is  con- 
structed with  a  radius  of  unit  length.     Draw  the  vertical  diameter 
BD  and  the  horizontal  diameter  FO.    Extend  FO  to  the  right 


HARMONIC  MOTION 


403 


of  the  circle  so  as  to  represent  a  horizontal  axis.  Construct  a 
vertical  axis  through  the  origin  0.  Locate  E  on  the  horizontal 
axis  so  that  OE  equals  the  rectified  arc  of  the  circumference. 
This  may  be  done  by  calculation  making  OE=2tzR.  Therefore 
OE  =  6.2832  (R  =  l)  units.  It  may  also  be  laid  off  approximately 
by  the  following  construction.     Divide  the  circumference  into  any 


convenient  number  of  equal  divisions,  say  24,  i.e.,  at  every 


12 


radian  or  15°  of  arc.  Beginning  at  0  number  these  divisions 
consecutively  from  0  to  24.  Lay  off  24  corresponding  con- 
secutive divisions  to  the  right  of  0,  each  being  equal  in  length 
to  the  chords  subtended  by  the  divisions  on  the  circle.  Number 
these  consecutively  from  0  to  24.     See  Fig.  (161). 


Fig.  161. — The  Sine  Curve  or  Record  of  Simple  Harmonic  Motion. 

Through  the  points  of  division  on  the  horizontal  axis  con- 
struct vertical  lines  and  through  the  arc  divisions  of  the  circle 
draw  lines  parallel  to  the  horizontal  axis.  These  sets  of  lines 
intersect.  Like  numbered  lines  determine  points  of  intersection 
through  which  we  pass  a  smooth  curve  which  is  sinusoidal.  This 
sine  curve  portrays  harmonic  motion  continuously.  The  complete 
cycle  is  represented  in  the  interval  OE. 

The  curve  is  identical  with  the  graph  plotted  from  Ex.  8, 
Chapter  XIV. 

3.  Plotting  the  Fundamental  Sine  Curve.  The  sine  curve 
may  be  plotted  from  (2)  which  reduces  to  (3)  when  R  =  l. 


(3) 


y  =  sin  6. 


(3)  is  the  equation  of  the  fundamental  sine  curve  to 
which  all  other  sine  curves  are  referred  for  comparison; 
(3)  is  plotted  from  Table  XXIV,  which  has  been  prepared 


404 


PRACTICAL  MATHEMATICS 


by  assigning  values  to  0  varying  in  gradations  of  30°  from 
0°  to  360°.  The  sines  of  multiples  of  30°  are  the  familiar 
values  0,  .5,  .866,  and  1,  and  should  be  selected  so  as  to 
minimize  the  labor  of  computation  although  the  sines  are 
1  easily  read  from  a  sine  table  or  a  slide  rule. 

TABLE  XXIV.     THE  VARIATION  OF  THE  ORDINATE  AND 
THE  ANGLE  OF  THE  SINE   CURVE 


y 

0 
0 

.5 
30 

.866 
60 

1 
90 

.866 

.5 

0      -.5 

-.866 

-1. 

-.866 

-.5 

0 

degrees 

120, 

150 

180 

210 

240 

270 

300 

330 

360 

0=o)t 

n-radians 

0 

IT 

3~ 

7T 

~2 

3 

5* 

6 

IT 

7tt 
6 

4x 
3 

3*- 
2 

5ir 
3 

lbr 
6 

2a- 

unit- 
radians 

.52 

1.05 

1.57 

2.09 

3.13 

3.14 

3.66 

4.18 

4.71 

5.23 

5.75 

6.28 

The  values  of  y  and  0  are  then  plotted,  y  ranges  from 
+1  to  —1  and  0  ranges  from  0°  to  360°,  i.e.,  from  0  to 
2iu  or  6.28  radians.  The  fundamental  sine  curve  is  shown 
as  curve  C  in  Fig.  163  and  as  y\  in  Fig.  165.  In  both 
figures  the  vertical  and  horizontal  scales  are  equal.  The 
horizontal  scale  may  be  graduated  in  degrees,  %  or  unit 
measure  corresponding  to  the  equivalent  values  of  0  in 
rows  2,  3,  4  of  Table  XXIV. 

Observation.  The  sine  curve  is  completed  in  360°  or  2%  or 
6.28  radians  and  consists  of  two  like  loops  or  arches  or  arcs 
which  are  alternately  placed  on  the  two  sides  of  the  horizontal 
axis.  The  two  loops  together  constitute  a  cycle  and  corre- 
spond to  a  cyclic  rotation  of  an  arm  which  moves  through 
360°  of  angle.  In  other  words,  0  has  changed  from  0  to  360°. 
while  y  has  increased  from  zero  to  +1,  decreased  from  -f-1 
to  zero  and  from  zero  to  —  1  and  increased  from  —1  to  zero. 
The  maximum  and  minimum  points  of  the  sine  curve  occur 
at  one-quarter  and  three-quarters  of  the  cycle.  The  zero 
points  occur  at  the  beginning,  at  the  half  cycle,  and  at  the 
end  of  the  cycle. 

4.  Angular  Velocity,  Period,  Frequency.  The  angular 
Velocity  (w)  of  a  rotating  or  revolving  body  is  numerically 


HARMONIC  MOTION  405 

equal  to  the  angle  or  arc  (0)  swept  through  in  a  unit  of  time 
(t)  as  expressed  in  (4),  (5),  and  (6). 

(4)  u>4 

(5)  e  =  w«, 

<•>  <4 

If  the  moving  arm  OP,  Fig.  160,  rotates  through  360° 
or  2iz  radians  in  a  unit  of  time,  i.e.,  it  makes  one  revolution 
per  second,  then  to  =2x.  If  the  arm  makes  25  revolutions 
per  second  then  to  =  25  X  2x  =  50z  =  50  X  3. 14  =  157.08.  If  the 
arm  makes  60  revolutions  per  second,  then  to  =  376.79.     If 

the  arm  makes  —  of  one  revolution  per  second  then  w  =  l. 

From  (5)  we  can  substitute  in  (3)  and  obtain  (7). 

(7)  y  =  sin  ut. 

The  interpretation  of  (7)  states  that  the  ordinate  of 
the  sine  curve  is  the  sine  of  the  product  wZ.  Therefore 
y  depends  upon  the  angular  velocity  of  the  rotating  arm 
and  the  time  in  seconds  which  has  elapsed  since  the  arm 
CP,  Fig.  160,  occupied  its  initial  position  CO.  In  (7)  w 
is  constant  and  t  is  variable,  therefore  in  plotting  (7)  the 
horizontal  axis  will  be  graduated  in  units  of  time. 

For  one  complete  cycle  6  attains  the  value  2x  and  therefore 
for  one  cycle  wZ  ranges  from  0  to  2tu.  By  substituting 
2tu  for  9  in  (4)  and  (6)  we  obtain  (8)  and  (9)  respectively: 

(8)  *>=J, 

The  interpretation  of  (8)  states  that  the  angular  velocity 
of  the  arm  can  be  computed  by  dividing  2x  by  the  time 
T  required  for  a  complete  rotation  of  the  arm.     The  numeric 


406  PRACTICAL  MATHEMATICS 

value  of  to  is  the  coefficient  of  t  when  the  sine  equation 
is  given  in  the  form  (7). 

The  interpretation  of  (9)  states  that  the  time  T  of  a 
complete  rotation  of  the  arm  is  computed  by  dividing 
2x  by  to,  i.e.,  by  dividing  2x  by  the  coefficient  of  t  when 
the  sine  equation  is  given  in  the  form  (7).  The  time  T 
for  a  complete  rotation  of  the  arm  is  a  special  value  of 
t  corresponding  to  the  time  which  elapses  for  a  complete 
cycle  of  the  sine  curve  and  is  called  the  period  of  the  curve. 
The  horizontal  scale  for  the  sine  curve  may  be  expressed 
as  a  measure  of  time.  The  numeric  value  of  the  period 
of  the  curve  is  placed  at  the  2x  or  360°  mark,  i.e.,  at  the 
end  of  the  cycle  and  proportionate  values  of  t  are  placed 
at  the  intermediate  points  of  division. 

The  number  of  cycles  per  unit  of  time  is  called  the 
frequency  (/)  of  the  curve  and  is  numerically  the  reciprocal 
of  the  period  of  the  curve  as  expressed  in  (10). 

The  interpretation  of  (10)  states  that  the  frequency 
of  a  curve  is  numerically  the  same  as  the  number  of  rotations 
per  second  of  the  moving  arm,  i.e.,  the  ratio  of  the  angular 
velocity  to  to  2iu. 

Ex.  2.     Make  entries  in  the  blank  spaces  in  Table  XXV. 

5.  Comparison  of  Sine  Curves  of  Different  Frequency. 

From  Table  XXV  we  observe  that  (16)  y  =  sm2izt  has 
a  period  of  one  second  and  a  frequency  of  one  cycle  per 
second.  If  we  plot  this  curve  its  cyclic  length  will  be 
one  unit,  which  is  the  numeric  value  of  its  period.  Any 
convenient  distance  on  the  horizontal  scale  may  be  taken 
as  a  unit  of  time.  In  like  manner  (12)  will  have  a  cyclic 
length  equal  to  6.28  units,  which  is  the  numeric  value  of 
its  period.  (11)  will  have  a  cyclic  length  equal  4x  =  12.5G 
units. 


HARMONIC  MOTION 


407 


TABLE  XXV.    ANGULAR  VELOCITY,  PERIOD  AND 
FREQUENCY 


Equation. 

Angle  o>t. 

Angular  Ve- 
locity w. 

Period 
in  sec 

2i=T 

Frequency 
per  sec. 

—  =/ 

2*     J 

(11) 

(12) 
(13) 

(14) 

(15) 
(16) 
(17) 
(18) 
(19) 

y  =  sin  M 
y  =  sin  t 
y  —  sin  2t 

y  =  sin  Zt 

y  =  sin  rd 
y  =  sin  2id 
y  =  sin  50£ 

.5t 

t 

2t 

•      3Z 

•d 

2%t 

\20t 

0.5 
1.0 
2.0 

3.0 

4x 

2x 

X 

2x 
3 

.5 

1 

60 

(20) 

.04 

(21) 

lOOx 

Fig.   162  illustrates  the  graphs  of   (11),   (13),  and  (14). 
(13)   has  four  times  the  frequency  of   (11)   and  therefore 


Fig.  162. — Sine  Curves  of  Different  Frequency. 


408  PEACTICAL  MATHEMATICS 

there  will  be  two  completed  cycles  of  (13)  for  a  half  cycle 
of  (11).  (14)  has  six  times  the  frequency  of  (11)  and  there- 
fore there  will  be  three  completed  cycles  of  (14)  for  a  half 
cycle  of  (11).     to  is  called  the  frequency  factor. 

Observation.  The  frequency  of  a  sine  curve  varies  directly 
with  to,  i.e.,  as  the  coefficient  of  t  in  the  angle  to?,  whereas 
the  period  varies  inversely  as  o>. 

Ex.  3.  On  one  sheet  of  cross-section  paper  plot  the  following 
groups  of  curves:  (a)-(ll),  (12),  (13) >  (6)-(ll),  (12),  (14); 
(c)-(12),  (13),  (15);  (d)-(12),  (14),  (15);  (e)-(12),  (15),  (16); 
(/)-(14),  (15),  (16);  (<7)-(17),  (18),  (19);  (A)-(18),  (19),  (20); 
(i)-(19),  (20),  (21). 

6.  The  Amplitude  of  a  Sine  Curve.  The  graphs  in 
Fig.  162  have  an  equal  amplitude,  i.e.,  an  equal  vertical 
displacement  from  the  horizontal  axis.  Equations  (11)  to 
(21)  inclusive  are  special  forms  of  (7)  in  which  special 
values  have  been  assigned  to  w.  (7)  is  a  special  form  of 
(2)  in  which  R  =  \.  R  is  the  radius  of  the  generating  circle, 
i.e.,  the  length  of  the  rotating  arm.  Suppose  R^l,  then 
Eqs.  (11)  to  (21)  will  be  modified  by  the  introduction  of 
a  multiplier  R  preceding  sin  bit.  The  value  of  R  determines 
the  maximum  height  and  minimum  depth  of  the  sine  curve. 
Therefore,  R  is  the  factor  which  determines  the  amplitude 
of  a  curve  and  R  is  called  the  amplitude  factor. 

Fig.  163  represents  the  plotted  graphs  of  (22),  (23), 
(24),  and  (25). 


(22) 

y  =  sin  6, 

(23) 

2/  =  |sin6, 

(24) 

y  =  3  sin  5t, 

(25) 

y  =  sml--- 

TZt 

=  cos  -7-, 
4 

(22)  and  (23)  are  represented  as  C  and  D  respectively. 


HARMONIC  MOTION 


409 


The  period  of  both  (22)  and  (23)  is  6.28,  but  their 
amplitudes  are  1  and  .5  respectively.  Their  plotted  points 
are  given  in  Table  XXVI,  in  which  it  will  be  observed 
that  for  like  values  of  0  the  corresponding  ordinates  are 
in  the  ratio  of  2  :  1.  C  may  be  identified  with  (7)  and 
therefore  with  (12).  The  horizontal  scale  of  Fig.  163  may 
be  interpreted  as  radians  or  time,  since  C  has  a  period 
of  2x.     D  has  the  same  period  as  (12)  and  (22)  but  only 


3 

A=Y=SIN(f-^)=cos^ 
B=Y=3SIN  5t 
C=Y=SIN  -e- 
D=Y=^SIN-9- 

2 

MB 

L^ 

^JLo ^c 

____ 

IT 

180° 
^3.14 

27T 

360° 

6.29 

k 

^  i 

^s\ 

i       : 

0 

t 

i     i 

r^ 

^ 

f^ 

Z3    ; 

1  A 

-1 

-2 

-3 

Fig.  163. — Sine  Curves  of  Different  Amplitude. 


one-half  the  corresponding  amplitude.      The  period  of  B 

.    2x 

is  —  =  1.26  but  its  amplitude  is  3,  which  is  three  times  the 
o 

amplitude  of  C  and  six  times  the  amplitude  of  D.     Eq. 

A  may  be  expressed  as  a  sine  curve  in  which  the  angle  is 

Tit 

or  as  a  cosine  curve  in  which  the  angle  —  is  the 


(H) 

complement  to  f  --— —-  J     The  cosi 


cosine  curve  does  not  intersect 

the  Y  axis  at  the  origin,  but  the  beginning  of  its  cycle  is 
bcated  to  the  left  of  the   Y  axis.     The  maximum  value 


410 


PRACTICAL  MATHEMATICS 


"$ 

a 

© 

NO 

1 

-.866 
-1 

-.866 
-.5 

© 

CO 

CO 

IO         00         ^ 

CO 
CO 
00        io        0 

- 

CO 
1 

CO 

CO 

1 

-4.66 
-4 

-3.33 
-2.66 

CM 

1 

CO 

CO          CO 

H             ^             © 
1                  1 

+  .66 

1.33 

2. 

n|^ 

jsi« 

i 

1 

1         1         1         1 

H  l<M    nlco    fc  IcO      O 
1             1             1 

hIco  n|co  hI<n 

i 

& 

jS|o^lwe§|«^|w^|o 

H 

J5|co<^|co  hjw 

H|co  nIco    o 

05 

s> 

© 

»o 

2.598 
30 

2.598 
1.5 

e 

-1.5 
-2.598 
-3  0 

-2.598 
-1.5 
0 

- 

o 

*o 
o 

as       «*       Oi       co 

O             i-H             1-4             CM 

CM          CO          tJH          iO 

00 
CM 

CO 

CO          00          <M 

CO             CO             TtH 

r^       oo       OS 

CO         —i         CO 

9       io       25 

O          t-h          CM 

r1    ^    -■ 

"3 

o 

H|cO 

H|C0    HltMCll^iol^ 

H 

t>i<o4tl«?JSl«« 

l— (  1 

Q 

56 

o 

HO 

CM 

CO                         CO 

CO                        CO          iO 

t*<          »0          ^          CM 

o 

CO 

»o        CO 

CM          rt<         »0 

l'            l"            l' 

CO 

CO          IO 

"r!       °i       o 
1          1 

<» 

o 

H|cO 

HICO   Hlc^fNl^^l50 

N 

^|co^|co^h 

T-l    \ 

o 

a 

o 

iO 

CO                    CO 

00         ,_,          00         iO 

© 

CO 

5       _< 
»o       oo      y 

l'     l" 

CO 
CO 
00       io       0 

r   i' 

<u 

o 

H|cO 

N|C0   HlOl^lcO^IcO 

i— i 
CO 

N 

II 
o 

O 

oo 

^|CD^|C0^|W 

00 
CM 

CO 

Jsi^ico  i 

■"*        II 

o 

o 

CO 
CO 

HARMONIC  MOTION  411 

of  the  cosine  curve  D  is  numerically  the  same  as  its  intercept 
on  the  Y  axis. 


The  whole  angle,  which  is  designated  as  0 


' 5t'  (2 -t) 


in  C,  B,  and  A  respectively,  must  range  through  2x  radians, 

whereas  the  range  of  values  of  t  depend  upon  <o,  i.e.,  the 

coefficient  of  t  in  the  whole  angle.     At  the  completion  of 

2x 
the  cycle  of  B,  5/  =  2x  and  therefore  t  = —  =  1.256  sec.     At 

5 

the    completion    of    the    half    cycle,  52  =x    and   therefore 

t='—  =  .628.     In  the  same  manner  for  A  when  (— — —J  =0, 
5  \2      4  / 

%t     x 
then  by  transposition   and  dividing  we  obtain  —  =—  and 

4     2 

t=  +2.     This  statement  means  that  the  cosine  curve  begins 

its  positive  loop  at  2  units  to  the  left  of  the  Y  axis,  i.e., 

two  seconds  earlier  than  curves  B,  C,  and  D.     The  values 

of  t  for  B  and  A  in  Table  XXVI  are  determined  from  the 

corresponding  values  of  ht  and  \~— ~)  respectively. 

Observation.  The  cosine  curve  is  a  sine  curve  which  has 
been  displaced  along  the  horizontal  axis  an  amount  equal  to 
one-quarter  of  a  cycle. 

Ex.  4.     Make  entries  in  the  blank  spaces  in  Table  XXVII. 

Ex.  5.  Fig.  (164)  illustrates  the  graphs  of  (34)  (35)  and  (36). 
The  vertical  scale  is  common  to  the  three  graphs.  The  horizontal 
scale  of  (36)  is  indicated  in  x  measure.  Determine  the  horizontal 
scales  for  (34)  and  (35). 

Observation.  The  horizontal  and  vertical  scales  of  sine 
curves  may  be  made  unequal  unless  the  true  shape  of  the  curve 
is  desired.  Sine  curves  of  different  frequency  may  be  plotted 
with  different  horizontal  scales  if  they  are  to  be  considered 
independently. 


412 


PRACTICAL  MATHEMATICS 


TABLE  XXVII.     AMPLITUDE 

AND  FREQUENCY 

No. 

Equation. 

Period. 

Frequency. 

Amplitude. 

(26) 

y  —  sin  2x2 

1 

1 

1 

(27) 

(28) 
(29) 
(30) 
(31) 

(fe) 

y=2sm  — 

y  —  1.5  sin  50x* 
i/  =  140  sin  .2* 
y  =  1050  sin  4t 
y=  —sm2%t 

-   •       * 
y  =  7  sm-— 

8 

2 

25 

1  5 

1050 

1 

(33) 

.       id 
2/=-sin-— 

(34) 

2/ =  2  cos  .7854* 

(35) 

y  =  .Q  sin  — 

(36) 
(37) 
(38) 
(39) 
(40) 

y  =  2  sin  1.5* 

60 

100 

15 

110 

10 

500 

.0005 

50 

Fig.  164. — Sine  Curves  of  Different  Amplitudes. 


HARMONIC  MOTION  413 

The  scale  of  a  sine  curve  is  determined  from  its  period. 
The  cyclic  length,  i.e.,  period,  is  most  conveniently  divided 
into  the  four  one-quarter  cycle  divisions  and  these  are  sub- 
divided into  thirds.  The  ordinates  at  the  thirteen  points 
including  the  initial  and  the  terminal  points  are  respectively: 

0,  .5,  .866,  1,  .866,  .5,  0,  -.5,  -.866,  -1,  -.866,  -.5,  0. 

When  the  curve  has  an  amplitude  differing  from  1  the 
above  ordinates  are  multiplied  by  the  amplitude. 

Ex.  6.  Plot  the  following  groups  of  curves,  using  the  same 
horizontal  and  vertical  scales:  (o)-(26),  (27),  (28);  (6)-(29), 
(37);  (c)-(31),  (33);  (rf)-(27),  (34),  (36);  (e)-(34),  (35),  (36); 
(/)-(30),  (39);  (flf)-(38),  (40);  (h)-(S2),  (38). 

7.  In  practice  it  is  customary  to  operate  electrical 
machinery  and  regulate  transmission  circuits  with  alter- 
nating currents  of  standard  frequency.  This  means  that 
the  alternating  stresses  in  the  circuit  or  net  work  are 
representable  by  sine  curves  of  equal  frequency  in  which 
y  is  replaced  by  e  or  i  the  instantaneous  values  of  the 
E.M.F.  and  current  respectively. 

There  is  another  factor,  however,  which  is  of  considerable 
importance  in  commercial  operation.  Two  sine  curves  may 
have  equal  frequencies  yet  when  plotted  they  will  be  rel- 
atively displaced  on  the  horizontal  direction.  When  corre- 
sponding zero  and  maximum  points  are  not  coincident 
but  are  separated  by  a  constant  interval  then  the  displaced 
curves  are  said  to  be  out  of  phase.  The  constant  interval 
of  separation  is  measured  in  time  or  when  measured  in 
radians,  or  degrees  it  is  called  the  phase  angle.  Fig.  165 
represents  two  arms  CP  and  CQ  which  rotate  with  like 
angular  velocity  and  are  constantly  separated  by  the  angle 
a.  The  projections  of  their  respective  extremities  P  and 
Q  give  harmonic  motions  which  are  represented  by  the 
respective   sine   curves   y\    and   y^   which   are   formulated 


414 


PRACTICAL  MATHEMATICS 


in  (41)  and  (42).     CP  and  CQ  are  designated  by  Ri  and 
R2  respectively. 


(41) 

(42) 


yi  =  Ri  sin  0, 
1/2  =  R2  sin  (J>. 


When  CP  is  in  the  initial  position  CO,  its  projection 
is  zero,  but  at  that  instant  CQ  is  in  advance  of  its  initial 
position  by  the  angle  a.  Therefore  the  sine  curve  2/2  is 
advanced  in  its  development  before  the  beginning  of 
2/i,  i.e.,  when  the  ordinate  of  y\  equals  zero  the  ordinate 
of    2/2    equals    R2  sin  a.     Maximum,    minimum,    and    zero 


Fig.  165. — Sine  Curves  Out  of  Phase. 

points  are  displaced  by  a  constant  distance  representing 
the  scaled  value  of  a.  The  angles  a,  4>,  and  0  are  related 
as  expressed  in  (43a)  and  (436).  Therefore  (41)  and  (42) 
are  rewritten  as  (44)  and  (45)  respectively  by  substituting 
from  (43a)  and  (436). 


(43a) 

<]>  =  0+a, 

(436) 

0  =  cl>-a, 

(44) 

yi=Ri  sin  ($  — a), 

(45) 

t/2  =  #2  sin  (0+a). 

Since  the  unequal  arms  CP  and  CQ  rotate  with  like 
angular  velocity,  both  corresponding  sine  curves  will  have 


HARMONIC  MOTION 


415 


a  like  frequency  and  therefore  they  are  represented  in 
Fig.  165  by  2/1  and  2/2,  which  have  an  equal  period  but 
unequal  amplitudes. 

Comparing  Eqs.  (41)  and  (42)  we  observe  that  the 
coefficients  of  0  and  cj>  are  unity  for  both  of  these  sine  curves, 
which  indicates  an  equal  frequency  and  an  equal  period 
for  2/1  and  2/2.  The  displacement  of  curve  2/2  in  advance 
of  2/1  is  indicated  by  the  added  angle  a  in  the  total  angle 
(8+ a)  of  (45)  and  therefore  2/2  is  said  to  be  out  of 
phase  with  2/1  or  2/2  leads  2/1  by  the  angle  a. 

Comparing  Eqs.  (44)  and  (45)  we  observe  that  the 
coefficients  of  §  and  0  are  unity  for  both  of  the  sine  curves, 
which  indicates  an  equal  frequency  and  an  equal  period  for 
2/2  and  2/1.  The  displacement  of  curve  2/1  in  the  rear  of  2/2 
is  indicated  by  the  subtracted  angle  a  in  the  total  angle 
(({>  —  a)  of  (44)  and  therefore  2/1  is  said  to  be  out  of  phase 
with  2/2  or  2/1  lags  behind  2/2  by  the  angle  a. 

In  the  comparison  of  the  graphs  (41)  and  (45)  we  observe 
that  when  2/1  =0,  then  sin  0  =  0,  since  R\  is  constant,  but 
when  sin  0  =  0,  then  0  =  0,  x,  2x  .  .  .  nx  where  n  is  an 
integer.  Accordingly  (44)  reduces  to  the  following  sim- 
ultaneous values: 


TABLE  XXVIII.     SIMULTANEOUS  VALUES  OF  yx  AND  y2  IN 

(41)   AND   (44) 


2/1 

e 

2/2 

0 
0 
0 

0 

X 

2x 

R2  sin  (+<*)  =  +R2  sin  a 
R2  sin  (x+a)  =  —R2  sin  a 
R2  sin  (2x+a)  =  +R2  sin  a 

The  displacement  between  2/1  and  2/2  is  constantly  equal 
to  a  as  shown  by  the  above  tabulated  values.  If  a  com- 
parison be  made  between  the  sine  curves  2/1  and  2/2,  regarding 
them  as  graphs  of  (42)  and  (44),  we  observe  a  like  dis- 
placement for  every  set  of  pairs  of  simultaneous  points. 


416  PKACTICAL  MATHEMATICS 

The  axal  displacement  of  two  sine  curves  of  period  T 
may  be  estimated  as  a  decimal  part  of  the  period  of  either 
curve. 

If  a  represents  the  angle  of  lead  or  lag,  then  —  represents 

2x 

the  displacement  as  a  deciaml  part  of  the  length  of  the 

a 

curve  and  —  T  is  the  unit  measure  of  the  displacement 

2x 

a 

which  reduces  to  —  as  shown  in  (46). 
(0 

(46)       ^-T  =  = — -  =  —  =  unit  measure  of  displacement. 
2x       2x  a)      a) 

The  interpretation  of  (46)  states  that  the  unit  displace- 
ment of  two  sine  curves  which  are  out  of  phase  may  be 
obtained  by  dividing  the  angle  of  lead  or  lag  by  the 
angular  velocity  w. 

(47a)        2/1  =  115  sin  50*, 

(476)        2/2  =  115sin  (50*+30°)  =  115  sin  (50*+|V 

In  the  plotting  of  (47a)  and  (476)  the  amplitudes  =  115, 

the  periods -^— =  .1257  units,   and    the  displacement =^7^ 
l ou  oUU 

.0105  units. 


12 

Ex.  7.  Group  the  following  equations  for  the  purpose  of 
comparing  their  leads  and  lags  and  then  tabulate  their  amplitudes, 
phase  angle,  frequency,  period,  and  displacement.  In  each  group 
assume  one  of  the  e  curves  as  a  standard  for  comparison. 

(48)  ex  =115  sin  (120x*+30°), 

(49)  ii  =50  sin  H20irf+~V 

(50)  ;2=25sin(120x*-.25), 

(51)  e2  =  120sin(120x<+.15), 


HARMONIC  MOTION  417 


(52)  e3  =  118sinl20x*, 

(53)  e4  =  115  sin  (50x^+30°), 


(54)  i3=25  sin 


(«*-§), 


(55)  u  =  50  sin  (50x*  +  .25) , 

(56)  eb  =  120  sin  (120x*  +45°), 

(57)  e6  =  120  sin  (50x*+45°), 

(58)  e7=220sin(50x*-.15), 

(59)  i5  =  118sin50x<. 

Ex.  8.  Plot  the  following  groups  of  equations  on  the  same 
cross-section    paper:     (a)-(52),    (48);     (6)-(52),    (51);     (c)-(52), 


(50) 
(51) 
(54) 
(54) 
(53) 
(56) 


(d)-(52),    (49);     (e)-(52),    (56);     (/)-(48),    (56);  (^)-(56), 

(A)-(56),    (49);     (i)-(56),    (50);     (j)-(59),    (53);  (*)-(59) 

(0-(59),   (55);    (m)-(59),    (57);    (n)-(59),   (58);  (o)-(58), 

(p)-(58),    (53);     fo)-(58),    (55);     (r)-(58),    (57);  (s)-(48), 

(0-(49),    (54);    (u)-(50),    (55);     (v)-(51),   (58);  (w)-(48)f 
(.r)-56),  (57);  fo)-(48)f  (57);  (z)-(48),  (59). 


8.  The  Resultant    of  Two   Simple  Harmonic  Motions. 

Two  simple  harmonic  motions  may  be  united  by  addition 
into  a  simple  harmonic  motion  which  is  called  their 
resultant.  The  resultant  will  have  a  period  equal  to  that 
of  its  components.  Fig.  166  represents  two  radii  Ri  and 
#2  which  move  uniformly  with  equal  periods  about  the 
center  C.  Their  respective  harmonic  formulas  are  expressed 
in  (60)  and  (61). 

(60)  yi  =  fiisin(w«+6i), 

(61)  2/2  =  #2  sin  M+ 62). 

The  two  distinct  motions  of  S  and  K  the  respective 
extremities  of  Ri  and  R2,  contribute  simple  harmonic 
motions  along  the  vertical  diameter.  These  displacements 
are  denoted  in  the  usual  manner  by  1/1  and  t/2  respectively. 
The  addition  of  y\  and  1/2  equals  y  the  displacement  of  the 


418 


PRACTICAL  MATHEMATICS 


projection  of  a  point  E  at  the  extremity  of  the  arm  B, 
and  therefore  (62),  (63),  and  (64)  follow: 

(62)  y  =  Vl +y2, 

(63)  y  =  Ri  sin  (ut+bi)+R2  sin  (w*+62), 

(64)  y  =  R  sin  (ut+b). 

Each  harmonic  curve  is  representable  by  a  sine  curve 
and  therefore  the  sum  of  the  two  simultaneous  ordinates 
of  curves  y\  and  y2  give  the  corresponding  ordinate  of  y. 


Fig.  166. — The  Resultant  of  Two  Simple  Harmonic  Motions. 

Observation.  Two  or  more  sine  curves  of  equal  period 
are  replaced  by  a  single  sine  wave  of  like  period.  The 
resultant  curve  has  its  ordinates  everywhere  equal  to  the  algebraic 
sum  of  the  ordinates  at  simultaneous  points  on  the  component 
curves. 

9.  The  resultant  B  in  Fig.  166  is  the  diagonal  of  a 
parallelogram  whose  adjacent  sides  are  numerically  equal 
and  parallel  to  Ri  and  R2.  By  the  law  of  the  oblique 
triangle  we  write  (65) : 


(65) 
but 
(66) 
(67) 
and 
(68) 


R  =  Ri2+R22+2RiR2  cos  CSE, 

CSE  =  %-(bi-b2), 

.     R  =  Ri2+R22+2RiR2  cos  (ic-fci+fo), 

R  =  Ri2+R22+2R!R2  cos  (6i-62). 


HARMONIC  MOTION  419 

The  interpretation  of  (68)  gives  the  working  formula 
for  computing  the  amplitude  of  the  resultant  in  terms  of 
the  amplitudes  of  the  components  and  their  phase  difference. 

The  sum  of  the  vertical  projections  of  Ri  and  R2  equals 
the  vertical  projection  of  R  and  the  sum  of  the  horizontal 
projections  of  Ri  and  R2  equals  the  horizontal  projection 
of  R.  The  ratio  of  these  sums  of  projections  equals  the 
tangent  of  the  phase  of  the  resultant  as  expressed  in  (69). 

,aCk,  ,       ,     Ri  sin  61 +R2  sin  b2 

(69)  tan  b  =  -=- ,     ,  p =-. 

Ri  cos  61 -\-R2  cos  02 

Ex.  9.     Add  the  ordinates  of  the  sine  curves  corresponding  to 
the  following  groups  of  equations:    (a)-(48),  (51);    (6)-(48),  (52) 
(c)-(48),    (56);     (d)-(51),    (52);     (e)-(5l),    (56);     (/)-(52),    (56) 
(<7)-(49),    (50);     (A)-(50),    (51);     (i)-(53),    (57);     (j)-(53),    (58) 
(*)-(57),  (58);  (Z)-(54),  (55);  (ro)-(54),  (59);  (n)-(55),  (59). 

10.  Periodic  motions  of  unequal  period  may  be  united 
to  produce  a  resultant  periodic  motion,  but  the  latter  does 
not  follow  the  law  of  simple  harmonic  motion. 

If  the  component  harmonic  motions  are  plotted  as 
sine  curves  then  the  sum  of  any  two  simultaneous  ordinates 
of  the  components  is  numerically  equal  to  the  simultaneous 
ordinate  of  the  resultant  curve. 

If  two  component  curves  are  represented  by  (70)  and 
(71)  then  their  resultant  is  expressed  by  (72)  and  (73). 

(70)  yi=Ri  sin  (8+&1), 

(71)  2/2  =  ^2  sin  (m0 +62), 

(72)  y  =  yi +y2, 

(73)  y  =  Ri  sin  (8+&i)+#2  sin  (?>iB+&2, 

Fig.  167  shows  the  component  curves  yi  and  y2  and 
their  resultant  yz. 

Observation.  The  period  of  the  resultant  of  two  or  more 
sine  curves  is  numerically  equal  to  the  period  of  the  curve  of 
least  frequency. 


420 


PRACTICAL  MATHEMATICS 


HARMONIC  MOTION  421 

11.  Fig.  168  shows  the  effect  of  uniting  two  sine  curves 
whose  relative  amplitudes  and  periods  are  both  in  the 
ratio  of  2:1.  The  effect  of  the  displacement  of  the  (6) 
curve  is  shown  in  the  formation  of  the  hump  whose  shape 
is  modified  by  further  change  in  the  relative  displacement 
or  relative  amplitudes. 

Ex.  10.  Plot  the  component  curves  as  well  as  the  resultant 
for  the  following  equations: 

(74)  y  =  2smt+hsm3t, 

(75)  y-2  sin  (J +1.0123)  +£  sin  St 

(76)  y  =  sin  x  +  J  sin  3x  +  \  sin  5x  +\  sin  7x, 

(77)  y  =  100  sin  X  +50  sin  (3x  -40°), 

(78)  y  =  150+100  sin  x  +50  sin  (3x  -40°), 

(79)  y  =100  sin  2x*+60  sin(6%J- 1.571)  +10  sin  (10x*-3.14), 

(80)  ?/=§-§  sin  2z, 

(81)  y  =  100+100  sin x  +50  sin  (3x-75°), 

What  is  the  significance  of  the  constant  term  in  (78),  (80), 
and  (81)? 

(82)  %  =  140  cos  2t  -  80  sin  .% 

(83)  i  =  140  cos  (.2* +15°)  -80  sin  2t, 

(84)  i  =  14  cos  {2t  - 15°)  -8  sin  2t, 

(85)  e  =  200  cos  2t  +350  sin  .2*, 

(86)  e  =200  cos  (.22  +  15°)  +350  sin  2t, 

(87)  e  =200  cos  (.2*  - 15°)  +350  sin  2t, 

(88)  e  =200  cos  .22+350  sin  (.22+15°), 

(89)  e  =200  cos.  22+350  sin  (.21-15°), 

(90)  2/=sin.22+sin.62, 

(91)  t/=sin.22-sin.62, 

(92)  y  =  cos  120x2  +cos  360x2, 


422 


PRACTICAL  MATHEMATICS 


HARMONIC  MOTION 


423 


(93)  y  =  cos  120x2  -  cos  360x2, 

(94)  y  =  5  sin  . 2* +10  sin  .62, 

(95)  y  =5  cos  120xJ+15  cos  360x«, 

(96)  y  =  1.5  sin  .22  -2  sin  .62, 

(97)  y  =  25  sin  50x2  - 10  sin  1 50x2, 

(98)  y  =  25  sin  50x2  +25  sin  150x2, 

(99)  y  =25  cos  50x2  - 10  cos  150x2. 

12.  The  Product  of  Two  Sine  Curves.  The  product  of 
the  ordinates  of  two  sine  curves  equals  the  ordinate  of  a 
compound  or  product  curve  which  is  also  periodic.  In 
electrical  applications  the  two  constituent  curves  are  of 
the  same  frequency  and  their  product  is  a  curve  of  twice 

(I)  CURRENT  IN  PHASE  WITH  E.M.F. 


----_H_ __.-- 

Fig.  169.— Power  Curve. 

the  frequency.  If  one  of  the  constituent  curves  represents 
a  harmonically  varying  E.M.F.  and  the  other  constituent 
curve  represents  a  harmonically  varying  current  then  their 
product  is  a  harmonically  varying  power  curve.  In  Fig. 
169  curves  D  and  C  are  sine  curves  representing  the 
respective  E.M.F.  and  current  in  a  circuit  and  correspond 
to  Eqs.  (100)  and  (101).  Their  product  is  represented 
by  the  power  curve  E.  OK  is  the  axis  of  the  D  and  C 
curves.     The  power  curve  E  is  also  a  sine  curve  whose 


424  PKACTICAL  MATHEMATICS 

axis  lies  midway  between  EJ  and  OK  as  expressed  in  Eqs. 
(102)  and  (103).  E  is  a  maximum  point  of  the  power 
curve  and  occurs  simultaneously  with  the  maximum  values 
of  D  and  C  of  the  E.M.F.  and  current  curve  respectively. 
The  maximum  ordinate  of  the  D  curve  is  110  and  the 
maximum  ordinate  of  the  C  curve  is  44  and  therefore 
the  product  110X40  =  4400  which  is  the  maximum  ordinate 
of  the  E  curve.  The  ordinate  at  J  =  4400  =  ( - 1 10)  X  ( -  40) , 
i.e.,  the  product  of  the  minimum  values  (  —  110)  and  (  —  40) 
of  the  D  and  C  curves  respectively. 

(100)  e  =  110  sine, 

(101)  i  =  40  sinG, 

(102)  p  =  a  =  4400  sin2  e, 

(103)  p  =  4400(J -I  cos  26)  =2200-2200  cos  26. 

The  interpretation  of  (103)  states  that  the  power  curve 
is  a  cosine  curve  of  twice  the  frequency  of  (100)  and  (101) 
and  that  its  axis  is  located  2200  units  above  OK.  The 
cosine  curve  has  an  amplitude  of  2200,  which  is  one-half 
the  product  of  the  amplitudes  of  the  constituents,  and 
being  negative  in  sign  its  position  is  displaced  90°  behind 
a  sine  curve  of  like  frequency.  These  facts  are  verified 
in  Fig.  169.  All  of  the  ordinates  of  the  power  curve  extend 
above  the  line  OK  and  therefore  are  positive.  The  power 
although  periodic  is  not  alternating  but  pulsating. 

13.  In  Fig.  170  the  E.M.F.  and  current  curves  MSW 
and  LNTB  respectively  are  represented  out  of  phase  as 
expressed  in  Eqs.  (104)  and  (105).  The  power  curve  is 
represented  by  a  cosine  curve  LPQARUWB  corresponding 
to  (112).  The  maximum  values  of  the  three  curves  do  not 
occur  simultaneously. 


(104)  e  =  110sin6, 

(105)  i  =  50  sin  (o-g-), 


HARMONIC  MOTION 


425 


(106) 
(107) 

(108) 

(109) 

(HO) 

(111) 
(112) 


p  =  ei  =  5500  sin  0  sin  (  0 — ^  j , 


p  =  5500  sin  0  \  sin  0  cos  - —  cos  0  sin  — 


0 


p  =  5500  \  sin2  0  cos  ~  —  cos  0  sin  0  sin 
I  6 

ccrvA  f  1  —  cos  20        x     sin  20   .    x 
p  =  5500  \ jz cos — —  sin  - 


6     ' 


p  =  5500 


§  cos  ^-  — §  cos  20  cos  '-  —  \  sin  20  sin  % 


6  ' 


p  =  2750  cos  | -2750  cos  (20 ~\ 
p  =  2349 -2750  cos  ^20 -|Y 

(II)  CURRENT  30°OUT  OF  PHASE  WITH  E.M.F. 

U 


£1 — -?r 


~3 H 3 6 a     -*- 

Fig.  170. — Power  Curve. 


The  interpretation  of  (112)  states  that  the  power  curve 
is  a  cosine  curve  of  twice  the  frequency  of  (104)  and  (105) 
and  that  its  axis  is  located  2349  units  above  the  line  LB. 
The  cosine  curve  has  an  amplitude  of  2750,  which  is  one- 
half  the  product  of  the  amplitudes  of  the   constituents. 


It  is  displaced  90°-}-—,  i.e.,   120°  behind  a  sine  curve  of 
6 


426  PKACTICAL  MATHEMATICS 

like  frequency.  These  facts  are  verified  in  Fig,  170.  All  of 
the  ordinates  of  the  power  curve  do  not  extend  above  the 
line  LB  and  accordingly  the  shaded  areas  indicate  negative 
power.  The  power  curve  although  periodic  is  not  alternating 
but  is  intermittently  positive  and  negative.  The  useful 
power  delivered  to  the  circuit  is  the  difference  between 
the  positive  and  negative  loops  per  cycle. 

The  increase  of  the  phase  angle  between  the  E.M.F. 
and  current  decreases  the  disparity  between  the  positive 
and  negative  loops  of  the  power  curve.  When  the  phase 
angle  equals  90°  the  positive  and  negative  loops  of  the 
power  curve  are  equal,  and  then  the  power  curve  is  seen 
in  its  normal  position  as  a  sine  curve,  with  its  axis  common 
to  the  axes  of  the  E.M.F.  and  current  curves. 

Ex.  11.  Plot  (113)  y  =sin2  0  and  show  that  the  resulting  graph 
represents  a  power  curve. 

Ex.  12.  Plot  the  power  curves  from  the  following  groups 
of  constituent  curves  whose  numbers  are  given  in  Ex.  10: 


(a)-(82),  (85);  (6)-(83),  (85) 
(e)-(83),  (86);  (/)-(84),  (86) 
(i)-(84),    (87);     (j)-(82),    (88) 


(c)-(84),    (85);  (rf)-(82),  (86); 

(flf)-(82),    (87);  (ft)-(83),  (87); 

(A)-(83),    (88);  ©-(84),  (88); 
(m)-(82),  (89);  (n)-(83),  (89);  (o)-(84),  (89). 

Observation.  A  power  curve  may  be  constructed  as  a 
cosine  curve  whose  amplitude  is  J  the  product  of  the  amplitudes 
of  the  constituent  E.M.F.  and  current  curves.  The  axis  of 
the  power  curve  is  removed  above  the  axis  of  the  constituents 
by  an  amount  which  is  the  product  of  the  amplitudes  of  the 
constituents  times  the  cosine  of  angle  of  their  phase  difference. 
The  frequency  of  the  power  curve  is  twice  that  of  its  constituents. 
The  position  of  the  power  curve  is  located  behind  the  position 
of  a  like  sine  curve  of  equal  frequency  by  an  amount  corre- 
sponding to  90°  plus  the  phase  angle  of  the  constituents. 
The  power  curve  passes  through  the  zero  points  of  both  the 
constituent  curves.  The  disparity  of  the  positive  and  negative 
areas  of  the  power  curve  decreases  as  the  phase  angle  of  the 
constituents  increases. 


HARMONIC  MOTION 


427 


14.  The   Curve   of   Damped   or   Decaying   Oscillation. 

One  of  the  most  important  curves  which  is  met  with  in 
the  study  of  transient  phenomena,  such  as  the  transmission 
of  intelligence  through  space,  is  known  as  the  curve  of 
damped  or  decaying  oscillation  and  is  represented  in  Fig.  171. 


(114) 
(115) 
(116) 


y  =  Re-Ktam(at+Q), 
y2  =  R  sin  (g)^+0). 


428 


PRACTICAL  MATHEMATICS 


The  oscillating  curve  is  a  product  curve  whose  ordinates 
are  obtained  by  multiplying  the  simultaneous  ordinates 
2/i  and  2/2  of  an  exponential  and  a  sine  curve  respectively. 
The  value  of  y\  is  expressed  by  (115)  and  the  value  of 
2/2  is  expressed  by  (116).  e  is  the  Napierian  base  and  has 
a  negative  exponent. 

In  Fig.  171,  (a)  represents  the  exponential  curve  for 
the  special  value  K=\,  (b)  represents  the  sine  curve 
for  which  R  =  l,  0  =  0,  and  u  =  5.  In  all  the  curves  t 
represents  time  and  has  been  replaced  by  x.  The  product 
curve  (c)  has  the  same  frequency  as  (b)  but  its  amplitude 
diminishes  continuously.  In  a  very  few  cycles  the  (c)  curve 
flattens  out  so  as  to  be  scarcely  distinguishable  from  the  X 
axis.  For  practical  purposes  this  statement  would  be  in- 
terpreted to  indicate  a  cessation  of  the  decaying  current 
curve  although  theoretically  it  would  persist  indefinitely. 

Ex.  13.  Plot  (114),  (115),  (116)  on  the  same  sheet  of  paper 
checking  the  graphic  work  by  calculation.  Use  the  following 
sets  of  values  given  in  Table  XXIX. 


TABLE  XXIX.     CONSTANTS  FOR  DECAYING  OSCILLATING 

CURVE 

No. 

R 

K 

a 

e 

No. 

R 

K 

ti 

0 

a 

1 

.02 

0 

i 

.5 

.025 

50x 

0 

b 

1 

.02 

30° 

3 

.5 

.025 

50x 

30° 

c 

1 

.02 

60° 

k 

.5 

.025 

120x 

0 

d 

1 

.02 

90° 

I 

.5 

.025 

120x 

30° 

e 

2 

.02 

30° 

m 

5 

.02 

2000x 

0 

f 

2 

.02 

60° 

n 

5 

.01 

120x 

15° 

9 

1 

.01 

2x 

0 

V 

5 

.05 

120x 

30° 

h 

1 

.01 

2x 

30° 

9. 

.5 

.02 

50x 

15° 

The  construction  of  angles  from  their  functional  values 
is  illustrated  in  Figs.  172,  173  and  174  and  these  represent 
Eqs.  (117),  (118)  and  (119)  respectively. 

(117)  cos<t>  =  costan-1    Vs-f^— -    . 

I  "2+ "1 J 


HARMONIC  MOTION 


429 


In  Fifir.  172  construct  AC=P2+Pi,  BC=P2-Pi, 
BD  =  l  unit  =  DE.  Erect  GDI  BE  making  BG  =  GE  =  2 
units.     Then  GD  =  y/Z.     Erect  CF LBD,  then 


Therefore 


CF-V3(Pi-Pi). 


Fig.  174. 


Verify   the    constructions   of    Figs.  173  and   174  from 
Eqs.  (118)  and  (119)  respectively: 


(118) 
(119) 


(j>  =  tan 


-Je  +  h 


mag 


<J>  =  sin 


-i_p 

Eh. 


CHAPTER  XXIII 
RATES,   DERIVATIVES,   AND  INTEGRALS 

1.  Speed.  We  refer  to  the  speed  of  a  train,  trolley-car, 
motor-car,  or  other  moving  body,  with  the  definite  idea 
that  a  precise  distance  has  been  traversed  in  a  specified 
time.  Speed  is  the  name  given  to  the  ratio  between  dis- 
tance and  time. 

A  train  which  traverses  a  distance  of  300  miles  in  6  hours 
has  an  average  speed  =  -£-  =  50  miles  per  hour.  At  the 
same  rate,  i.e.,  speed,  the  train  would  travel  20  miles  in 
24  minutes. 

Ex.  1.  What  is  the  distance  traversed  by  a  car  which  is  pro- 
pelled for  15  minutes  at  a  speed  of  30  miles  per  hour? 

Ex.  2.  Construct  a  speed  chart  from  the  following  time-table 
XXX.  Plot  distances  vertically  and  time  horizontally.  The 
horizontal  axis  should  provide  for  24  hours. 

Ex.  3.  Determine  the  average  rate  of  travel  for  each  train  by 
dividing  the  total  distance  by  the  total  time.  Calculate  the  rate 
of  travel  between  the  stations  at  which  the  train  stops  and  verify 
these  results  by  showing  that  the  slope  of  the  speed  curve  is  a 
measure  of  the  rate  of  travel  between  station  stops.  Draw  a  straight 
line  between  the  first  and  last  point  on  each  speed  curve  and  show 
that  its  slope  is  the  average  rate  of  travel  for  the  entire  distance. 
The  following  train  numbers  are  for  investigation:  (a)  51;  (6)  59; 
(c)  63;  (d)  65;  (e)  67;  (/)  69;  (g)  71;  (h)  73;  (i)  75;  (j)  77; 
(k)  79;  (0  81;  (m)  85;  (n)  89;  (o)  91;  (p)  93. 

2.  Velocity.  The  word  velocity  is  used  often  synono- 
mously  for  speed.  Strictly  interpreted  velocity  involves 
the  notion  of  direction  as  well  as  speed.  In  the  discussion 
of  the  velocity  Of  a  moving  body  it  is  necessary  to  consider 

430 


RATES,  DERIVATIVES,  AND  INTEGRALS 


431 


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432 


PEACTICAL  MATHEMATICS 


the  uniformity  or  non-uniformity  of  the  speed,  i.e.,  the 
constancy  or  variability  of  the  quantity  of  motion  in  a 
unit  of  time.  A  unit  of  time  may  be  an  hour,  a  minute,  or 
a  second  or  a  lesser  interval  which  depends  upon  the  nature 
of  the  problem  and  the  refinements  of  measurement  and 
observation. 

We  form  a  very  good  understanding  of  these  facts  by 
considering  the  vicissitudes  of  a  journey  between  New 
York  and  Chicago.  Straight,  level  roadbeds  are  inter- 
spersed by  circuitous  winding  stretches  of  track.  The 
ascent  of  an  incline  in  one  interval  is  followed  by  a  descent 


Chicago 


CD  10 

N«w  York  Time  in  Hours 

Fig.  175.— Chart  of  Variable  Speed. 

of  a  decline  in  the  next  interval.  In  consecutive  inter- 
vals of  time  both  variable  and  constant  velocities  will  be 
observed.  Such  a  record  of  performances  is  shown  in 
Fig.  175.  Time  is  plotted  horizontally  and  distance  verti- 
cally, so  that  the  slope  of  the  heavy  black  line  OBEMQVX 
is  the  measure  of  the  speed  of  the  train,  i.e.,  its  rate  per 
hour.  At  every  instant  the  speed  is  changing  and  it  is  by. 
considering  short  lapses  of  time  that  we  can  approximate 
to  the  actual  rate  at  any  instant. 

3.  Slope.     The  slope  of  a  curve  is  measured  by  the 
slope  of  the  lines  which  are  tangent  to  the  curve.     The 


RATES,  DERIVATIVES,  AND  INTEGRALS         433 

slope  of  a  line  is  proportional  to  the  tangent  of  its  angle  of 
inclination  with  the  horizontal.  The  slope  of  the  curve 
OBEMVX  at  the  respective  points  B,  E,  M,  V,  and  X  is 
numerically  the  same  as  the  respective  slopes  of  the  tan- 
gents GB,  HE,  LM,  SV,  and  WX.    Therefore  the  rate  of 

the  train  at  B  =  R  tan  6=77^.     The  slope  at  E  =  R  tan  A. 

R  is  the  ratio  of  the  vertical  to  the  horizontal  scale. 

Ex.  4.    Show  that  the  slopes  at  M,  Q,  V,  and  X  in  Fig.  175  are 

expressed  by  the  respective  ratios:  -rrr,  777;,  7777,,  and  77777. 


Fig.  176.- 


C  D 

-Initial,  Average,  and  Final  Velocities. 


4.  The  Scale  of  Velocity.  In  Fig.  176  the  tan  0  measures 
the  velocity  of  the  train  as  it  passes  station  B,  whereas  the 
tan  cj)  measures  the  velocity  of  the  train  as  it  passes  station 
E.  These  rates  are  expressed  as  miles  per  hour,  because 
the  vertical  scale  is  in  miles  and  the  horizontal  scale  is  in 
hours  and  therefore  their  ratio  gives  a  rate  in  miles  per 
hour. 


434  PRACTICAL  MATHEMATICS 

5.  Initial,  Average,  and  Final  Velocities.  If  the  velocity 
is  variable  during  any  increment,  i.e.,  change  of  time,  then 
the  average  velocity  is  the  ratio  of  the  increment  of  distance 
to  the  increment  of  time.  In  the  speed  chart,  Fig.  176, 
BF  —  CD  =  50  min.  is  an  increment  of  time.  The  corre- 
sponding increment  of  distance  =EF=ED—FD=ED—BC 
=  35  miles.  Therefore  the  average  velocity  (Va)  between 
B  and  E  is  expressed  in  (1)  and  (2).  An  increment  is  ab- 
breviated by  A,  so  that  At  is  an  abbreviation  for  the  in- 
crement or  interval  of  time  and  As  is  an  abbreviation  for 
the  increment  of  distance  or  space  traversed. 

v      increment  of  distance  _EF 
increment  of  time        BF' 

(2)  va  =  ^  =  R  tan  EBF. 

The  initial  velocity  (Vt)  is  the  velocity  at  B,  i.e.,  at 
the  beginning  of  the  time  interval,  whereas  the  final  velocity 
(V/)  is  the  velocity  at  E,  i.e.,  at  the  end  of  the  time  interval. 

(3)  F«=#tan6.  (4)  Vf=Rtsai^. 

(5)  Vi>Va>Vf. 

(6)  /.     tan  6  >  tan  EBF>  tan  <i>. 

The  interpretation  of  (5)  and  (6)  states  that  average 
velocity  is  less  than  the  initial  velocity  but  greater  than 
the  final  velocity. 

(7)  /.      Vi-M>Va-M>VrM. 

(7)  results  from  (5)  by  multiplying  the  inequality  (5)  by 
At.  The  interpretation  of  (7)  states  that  the  actual  incre- 
ment of  space  which  equals  EF  =  As  =  VaAt  is  less  than 
the  increment  which  would  be  gained  were  the  initial 
velocity  Vt  to  continue  during  a  like  interval  At,  and  is 
greater  than  the  increment  which  would  be  gained  were 
the  final  velocity  V/  to  continue  during  a  like  interval  At. 


EATES,  DERIVATIVES,  AND  INTEGRALS         435 

In  the  above  illustration  the  increment  of  time  is 
50  minutes;  in  other  words,  the  station  E  is  reached  50 
minutes  after  B  and  the  velocity  has  changed  correspond- 
ingly from  Vi  to  Vf.  If  stations  are  noted  successively 
closer  to  B  the  corresponding  time  interval  will  decrease 
and  the  average  velocity  will  approach  its  limiting  value, 
which  is  the  initial  velocity.  If  the  slope  corresponding  to 
the  successive  average  velocities  be  designated  by  tan  a, 
then  (9)  follows  from  (8). 


(8) 

The  limit  of  Fa  =  7«. 

The  limit  of     a  =  6. 

(9) 

The  limit  of  tan  a  =  tan  0, 

but  the 

■s 

(10) 

.             As 
tan  a  =  — . 
At 

(11) 

MzdO  At 

At^J)  means  as  At  approaches  (=)  zero.     The  symbol 

ds 

~r  is  an  adopted  abbreviation  for  the  left-hand  member 

of  (11)  and  is  known  variously  as  a  derivative  of  space, 
differential  coefficient  of  space,  or  a  time  rate  of  change  of 
space.     Substituting  the  abbreviation  in  (11)  we  obtain  (12). 

ds 
(12)  -7-  =  tan  0  =  instantaneous  velocity  =  V. 

The  interpretation  of  (12)  states  that  V,  the  velocity  at 
any  instant,  is  the  limiting  value  of  the  average  speed  meas- 
ured to  or  from  that  instant.  In  other  words,  we  approach 
more  nearly  to  the  value  of  the  velocity  of  a  train  at  each 
successive  instant  as  we  shorten  the  time  interval  which 
elapses  during  an  observation.  Hence  arises  the  definition 
that  velocity  is  the  time  rate  of  change  of  space  or  distance. 


436  PEACTICAL  MATHEMATICS 

These  facts  which  we  have  observed  from  a  record  of  the 
speed  of  a  moving  train  are  applied  in  an  identical  manner 
to  the  graphical  records  of  every  changing  action  or  phe- 
nomenon. The  rate  of  change  of  any  plotted  action  or 
phenomenon  is  indicated  by  the  slope  of  its  tangent  lines. 

Observation.  When  a  body  moves  uniformly  then  in  equal 
consecutive  increments  of  time  it  traverses  equal  distances.  This 
fact  is  illustrated  by  a  graph  for  which  the  coordinate  axes 
correspond  to  distance  and  time.  Linear  graphs  are  the  only 
graphs  which  have  a  constant  slope  and  therefore  uniform 
motion  is  represented  graphically  by  straight  lines  and  non- 
uniform motion  is  represented  by  non-linear  graphs. 

6.  Linear  and  Angular  Velocity.  When  an  electric 
motor  maintains  a  constant  speed  then  every  conductor 
on  the  periphery  of  the  armature  has  a  constant  linear 
velocity,  i.e.,  the  distance  or  length  of  arc  traversed  will 
be  equal  for  equal  intervals  of  time.  The  path  of  the  wire 
is  a  circle.  If  the  motor  maintains  a  constant  speed  of  n 
revolutions  per  second  then  the  linear  velocity  v  of  a  con- 
ductor is  the  length  of  the  circumference  (I)  times  the 
number  (n)  of  revolutions.  The  linear  velocity  of  a  rotating 
body  is  by  the  interpretation  of  (13)  the  rate  of  lengthening 
of  the  arc  per  second. 

(13)  »  =  ln  =  ft. 

The  coil  containing  the  conductor  generates  equal  angles 

in  equal  intervals  of  time.     In  the  same  circle  equal  arcs 

subtend  equal  central  angles.     The  rate  of  change  of  an 

angle  expressed  in  radians  is  called  angular  velocity  (w). 

dti 
If  the  angle  is  designated  as  0,  then  w  =  — -. 

dt 

(14)  1  =  2%R. 

(15)  v  =  2itRn  =  R2Kn. Sub  (14)  in  (13) 


RATES,  DERIVATIVES,  AND  INTEGRALS 


437 


In  (15)  2xn  is  the  radian  change  per  second  and  is  there- 
fore replaceable  by  <•>  as  shown  in  (16). 


(16) 


V  =  R<j)  =  R  -TT. 

at 


VOC(D. 


The  interpretation  of  (16)  states  that  the  linear  velocity 
of  a  rotating  body  is  directly  proportional  to  its  angular 
velocity  and  that  the  proportionality  factor  R  is  the  radius 
of  the  circular  path. 

Ex.  5.  Fill  in  the  blank  spaces  in  Table  XXXI.  n  is  expressed 
as  revolutions  per  minute,  whereas  v  and  &>  are  the  respective 
velocities  in  feet  and  radians  per  second.  R  is  the  radius  in 
inches. 


TABLE 

XXXI 

No. 

R 

n 

r 

CO 

No. 

R 

n 

r 

de 

dt 

No. 

R 

n 

dt 
dt 

de 

dt 

1 

10 
25 

36 
50 

50 

300 

80 
2x 

8 

200 

800 

15 

..       .. 

...    |    120x 

2 

7 
50 

60 
100 

9 

500 

16 

...    |      50x 

3 

10 

11 

77 

17 

81 

18 

...    |    ... 

4 

11 

62.8 

18 

12 

5000     .  .  . 

5 

12 

39 

19 

12 

650 

...    j    ... 

6 

13 

IOOtt 

20 

500 

3000     .  .  . 

7 

14 

42 

360 

21 

200 

2500     .  .  . 

Observation.  All  points  in  a  crank  arm  or  coil  rotate  with 
equal  angular  velocity,  but  their  respective  linear  velocities 
are  directly  proportional  to  the  radii  of  their  circular  paths. 

7.  The  Derivative.  Every  linear  formula  represents  a 
constant  or  uniform  ratio  of  change  between  two  variables, 
i.e.,  a  constant  rate  of  change  between  two  variables.  The 
slope  of  a  curve  for  any  of  its  points  is  the  ratio  between 
the  rates  of  change  of  two  variables  at  that  instant.  When 
one  of  the  variables  is  time,  the  slope  is  the  time  rate  of 
change  of  the  other  variable.  The  graph  NMTFP  in 
Fig.  177  represents  a  non-uniform  motion  and  has  been 
constructed  from  a  non-linear  formula.     At  each  successive 


438 


PRACTICAL  MATHEMATICS 


instant  the  rate  of  change  of  the  variables  is  different  from 
the  rate  at  the  preceding  and  following  instants.  At  N 
the  rate  of  change  is  measured  by  tan  a. 

rate  of  the  ordinate 


(17) 


(18) 


tan  a 


rate  of  the  abscissa' 


rate  of  y 

tan  a  =  —. — ~. 

rate  of  x 


S     A  BC  DEF  GH 

Fig.  177. — Representation  of  Non-uniform  Motion. 


The  symbol  for  abbreviating  the  right-hand  member  of 
(18)  is  -j-  and  hence  d  is  not  a  multiplier  but  a  symbol  of 

operation,     dx  is  an  abbreviation  for  the  rate  of  x,  i.e., 

dx 
dx  is  a  contraction  for  the  time  rate  -jr. 

at 


(19) 


tan  a 


dy        , 
dy _dt  _ time  rate  of  y 
dx     dx    time  rate  of  x' 

dt 


(20) 


tan  a  =  ^(2/)- 


RATES,  DERIVATIVES,  AND  INTEGRALS         439 

The  interpretation  of  (19)  states  that  the  slope  to  a 
curve  is  the  ratio  of  the  time  rate  of  change  of  the  ordinate 
to  the  time  rate  of  change  of  the  abscissa.     (20)  is  another 

form  of  (19)  in  which  -r-  shows  itself  more  distinctly  as  a 

symbol  of  operation  which  acts  upon  y.  The  d  in  the 
numerator  and  the  d  in  the  denominator  are  not  factors  or 
multipliers.  The  entire  symbol  made  of  the  two  d's  and 
the  x  is  an  abbreviation  for  a  definite  operation,  just  as 
log,  sin,  and  esc  are  three-lettered  abbreviations  for  definite 
operations. 


-.5 


Fig.  178. — The  Graphic  Determination  of  the  Derivative  of  a  Curve. 


8.  The  Graphic  Determination  of  the  Derivative  of  a 

Curve.     The  graphic  determination  of  the  derivative  of  a 

curve  is  the  operation  of  measuring  the  slope  of  a  curve 

at  its  successive  points.     In  Fig.  178  y\  is  a  sine  curve, 

whose   ordinates   are  instantaneous  values   of  current    (i) 

and  whose   abscissas  are  radians  and  whose  equation  is 

y  =  sm  0.     Tangents  have  been  drawn  at  the  lettered  points 

of  the  curve  and  their  measured  slopes  express  the  deriv- 

dv 
ative  of  the  curve  — ,  i.e.,  the  rate  of  change  of  the  ordinate  y 

with  the  angle  0,  according  to  Table  XXXII. 


440 


PRACTICAL  MATHEMATICS 
TABLE  XXXII 


Point  on  the 
curve 

0 

A 

B 

C 

D 

E 

F 

G 

H 

/ 

X 

L 

1/ 

Slope  of  tan- 
dy 

gent       'Te 

1 

.866 

.5 

0 

-.5 

-.866 

-i 

-.866 

-.5 

0 

.5 

.866 

1 

Ordinate     of 
the  curve,  y 

0 

.5 

.866 

1 

.866 

5 

0 

-.5 

-.866 

-1 

-.866 

-.5 

0 

Abscissa     of 
the  curve,  0 

0 

7T 

6 

7T 

3 

2" 

2x 
3 

5x 
6 

7T 

7tt 
6 

4tt 
3 

3tt 
2 

5r 

3 

11* 

6 

2x 

It  is  observed  that  the  numeric  values  of  slopes  follow  a 
cycle  of  changes  corresponding  to  the  ordinates  of  the  curve, 
excepting  that  the  initial  values  are  displaced.  The  suc- 
cessive values  of  the  derivatives  are  plotted  as  ordinates 
of  a  new  curve  labeled  1/2  which  is  also  a  sine  curve  but 
which  leads  y\  by  90°.     The  equation  for  y2  is  given  in  (21). 


(21) 
but 

(22) 
(23) 


2/2  =  sin  (6+90°)=  cos  6, 


d{y)      d  ,  .    a. 


d_ 

m 


(sin  0)=cos  6. 


The  interpretation  of  (23)  states  that  a  new  curve 
called  a  derivative  may  be  derived  from  a  given  curve 
which  is  called  its  primitive.  The  ordinates  of  the  deriv- 
ative curve  are  the  respective  rates  of  change  of  the  simul- 
taneous ordinates  of  the  primitive  curve.  (23)  also  states 
that  the  derivative  of  a  sine  is  a  cosine. 

9.  In  Fig.  179  y  =  Ismid  is  a  primitive  curve  and  its 
corresponding  derivative  is  the  curve  yt  =  to  J  cos  ut.  Both 
curves  have  a  like  frequency  but  the  amplitude  of  the 
derivative  is  w  times  the  amplitude  of  the  primitive.  The 
derivative  may  be  represented  by  the  dotted  cosine  curve 


RATES.  DERIVATIVES,  AND  INTEGRALS 


441 


in  Fig.  179,  in  which  case  its  scale  is  w  times  the  scale  of  the 
primitive. 


(24) 
(25) 


y  =  I  sin  bit. 

(y)  =yt  =  ul  cos  bit 


(26) 


dt 


(7  sin  bit)  =  b)I  cos  int. 


Y^wTcosOJf  , 

>1 


Fig.  179. — A  Primitive  and  its  Derivative. 


The  interpretation  of  (26)  states  that  the  derivative  of 
a  sine  curve  is  a  cosine  curve  and  that  the  ratio  of  the 
amplitude  of  the  derivative  to  the  amplitude  of  the  prim- 
itive is  2x  times  the  frequency  of  the  curves. 

Ex.  6.  Plot  y=x2  and  determine  its  slope  at  ten  distinct  points. 
Plot  the  slopes  as  ordinates  of  a  derivative  curve  which  should 
produce  a  linear  graph.  Determine  the  equation  of  the  derivative. 
The  tangents  may  be  most  readily  constructed  as  follows:  Draw 
a  chord  between  any  two  points.  Construct  a  perpendicular 
bisector  to  the  chord  and  where  the  perpendicular  intersects  the 
curve  draw  a  tangent  parallel  to  the  chord.  The  work  is  made 
easier  if  a  bisected  chord  of  fixed  length  is  moved  about  the  curve. 

Ex.  7.  Plot  the  derivative  curves  for  the  following  primitives: 
(a)  y=2x2;  (6)  y=2.5x*;  (c)  y  =  .2x*;  (d)  y=3x*;  (e)  y=2Ax>; 
(f)  y=2M\ 


442  PRACTICAL  MATHEMATICS 

Ex.  8.  Plot  the  derivative  curves  for  the  following  primitives : 
(a)  y=x3;(b)  y=2xz;  (c)  y  =  .5x3;  (d)  y=x*;  (e)  y  =  .8x*;  (/ ) 
y  =  .5x4. 

Ex.  9.  Construct  a  sine  curve  for  the  equation  i=I  sin  8  and 
determine  its  derivative  curve. 

Ex.  10.  Construct  the  primitive  and  its  derivative  for  the 
following  equations:  (a)  y=sm2t]  (b)  y  =  .5  sm2id;  (c)  i  = 
10  sin  50x£;  (d)  e  =  110  sin  120x*. 

Ex.  11.  Show  that  the  derivative  of  a  negative  sine  curve  is  a 
negative  cosine  curve. 

Ex.  12.  Show  that  the  derivative  of  a  negative  cosine  curve  is 
a  sine  curve. 

Ex.  13.  Show  that  the  derivative  of  a  cosine  curve  is  a  negative 
sine  curve. 

Ex.  14.  Plot  the  sum  of  the  (a)  equations  in  Exs.  7  and  8, 
and  show  that  the  derivative  of  the  sum  is  equal  to  the  sum  of 
the  derivatives. 

10.  Differentiation  of  Formulas.  The  differentiation  of 
formulas  is  the  mathematical  process  by  means  of  which  a 
primitive  equation  is  transformed  into  a  derivative  equa- 
tion. By  comparing  the  primitive  with  the  derivative  a 
formal  statement  or  law  of  differentiation  may  be  framed 
for  different  types  of  primitives.  The  primitive  (27)  trans- 
forms into  the  derivative  (28) . 

8      I  ■'%  1 

1       I    jjjj 

O         O     PM 

(27)  y  =  a  xn. 

(28)  ^anx"-1 

Comparing  (28)  with  (27)  we  observe  the  following  law 
for  writing  the  derivative  of  a  power  of  a  variable.  Mul- 
tiply the  coefficient  of  the  primitive  by  the  exponent  of  its 
variable  and  decrease  the  exponent  of  the  variable  by  unity. 

Ex.  15.  Verify  the  following  derivatives  by  applying  the  law 
of  a  power  to  the  primitives.  The  derivative  of  an  algebraic  sum 
equals  the  sum  of  the  derivatives  of  its  several  terms.    The  deriv- 


RATES,  DERIVATIVES,  AND  INTEGRALS        443 

ative  of  a  constant  term  is  zero.     Fill  in  the  blank  spaces  in  Table 
XXXIII. 

TABLE  XXXIII 


Primitive. 


Derivative. 


y=ax3 


dy 
dx 


Sax7 


y  =  ox 


dy 
dx 


:5z°=5 


ax* 
y  =  ~2 


dy 

~=ax 

dx 


y  =  5x*=5v/x3 

i 

^«7.5**-7.5vT 

dx 

y=ax-U25 

^=-1.25ax-2-25 
dx 

y=  —  5x~* 

dx     2X 

a            . 

y  =  -  =  ax  — 1 
u      X 

dy 
dx 

y  =  ax3-\-5x 

|-3«x.+5 

y  =  5xz+2x2+3x+5 

^  =  15x=+4,+3 

s  =  .5gt2+bt+c 

^S                        lit. 

Tt=v  =  gt+b 

y=x>+^ 

dy  = 
dx 

p  =  Rtv~l 

dp 

dv~ 

E  =  \ax2+d 

dE_ 
dx 

s  =  16.lt2+16.1t 

ds_ 
dt 

444 


PRACTICAL  MATHEMATICS 


Ex.  16.  Apply  the  law  of  differentiation  of  a  power  to  the 
primitives  given  in  Exs.  7  and  8.  Substitute  values  of  x  in  the 
derivative  equations  and  verify  the  simplified  numeric  results  by 
identification  with  the  ordinate  values  found  in  Exs.  7  and  8. 
This  substitution  is  shown  in  Table  XXXIV  for  the  primitive  (29) 
and  its  corresponding  derivative  (30). 

dy 


(29) 


y=2x\ 


(30) 


dx 


=4x. 


TABLE    XXXIV.— NUMERIC  VALUES    FOR   PRIMITIVE 
(29)   AND   DERIVATIVE   (30) 


Values  of  the  abscissa  x. . 

0 

.5 

1 

1.5 

2 

2.5 

3 

3.5 

4 

4.5 

5 

Values  of  the  ordinate  y  . 

0 

.5 

2 

4.5 

8 

13 

18 

24.5 

32 
16 

40.5 
18 

50 

Values  of  the  ordinate  -— 
dx 

0 

2 

4 

6 

8 

10 

12 

14 

20 

Observation.  The  derivative  of  an  algebraic  equation  is 
an  algebraic  equation  and  the  degree  of  the  derivative  is  one 
less  in  unit  value  than  the  degree  of  its  primitive. 

11.  The  Proof  of  the  Law  of  the  Derivative  of  a  Power. 
The  law  for  the  derivative  of  a  power  is  formally  proven  as 
follows:  Construct  the  graph  of  (31). 

(31)  y  =  axn. 

Select  any  point  P  on  the  curve,  construct  and  label  its 
ordinate  and  abscissa  y  and  x  respectively.  Join  P  with  a 
neighboring  point  Q.  Designate  and  label,  the  ordinate  and 
abscissa  of  Q  by  y-\-Ay  and  x-\-Ax  respectively.  The  incre- 
ments indicate  the  growths  or  increases,  i.e.,  differences 
between  the  respective  ordinates  and  respective  abscissas. 
By  substituting  the  coordinates  of  Q  in  (31),  we  obtain  (32). 

(32)  y+Ay  =  a(x+Ax)n. 

Join  P  with  Q.     The  line  PQ  forms  the  hypotenuse  of  a 

right  triangle  whose  vertical  side  is  Ay  and  horizontal  side 

is  Ax.     Suppose  Q  is  moved  indefinitely  near  P,  then  Ax 

approaches  zero,  PQ  approaches  the  position  of  the  tangent 

Ay  dy 

through  P  and  therefore  —  approaches  its  limiting  value  — -. 

-A.'-  ax 


RATES,  DERIVATIVES,  AND  INTEGRALS         445 

Subtracting  (31)  from  (32)  we  obtain  (33), 

(33)  Ay  =  a(x+Ax)n-axn=a{(x-\-Ax)n-xn}. 

The  parenthetical  quantity,  (x-{-Ax)n,  may  be  expanded 
by  means  of  the  Binomial  Theorem  given  on  page  240, 
and  (33)  becomes  (34)  which  simplifies  by  cancellation 
into  (35): 

(34)  Ay  =  a\(xn  +  nxn  lAx+n^^^Ax2 

H — ~^ Ax?  +  .  .  .+Axn)-xn 


(35) 


Ay  =  a\  nxn~1Ax-\ — - — ~- -Ax2 


— Ax?-{-.  .  .  Axtl 


3! 


(36)     S.     fx  =  a\\xn^  +  <n-VX'~2Ax+.  .  .  +A^-i 

(36)  results  from  dividing  (35)  by  Ax.    (36)  is  an  expression 

Ay 
between  the  two  always  equal  variables  —  -  and  the  terms 

of  the  right-hand  member.  By  the  Fundamental  Theorem 
on  limits  given  on  page  227  the  limits  of  the  two  members  of 

(36)  are  equal  and  therefore  (37),  (38),  and  (39)  follow  from 
(36).  Since  the  limit  of  Ax  =  0,  then  the  limit  of  every 
term  containing  Ax  is  zero.  The  constant  term  ax11'1 
remains  unaltered  in  the  operation  of  obtaining  limits. 

/0_    ,.      Aw     ..         f  .  .  n(n—l)x'l~2Ax  .  .   .   n_,  1 

(37)  hm— -  =  hmaj  nxn  l-\ — g, h-  •  •  +Aa;n  l    . 

(38)  .*.     -7-  =  anxn - 1+ zero + zero +.  .  .  zero. 

ax 

(39)  ...     c^  =  anxn-1. 

Therefore  to  write  the  derivative  of  a  power  of  a  variable 
multiply  the  exponent  into  the  constant  coefficient  and 
decrease  the  exponent  of  the  variable  by  unity. 


446  PRACTICAL  MATHEMATICS 

12.  The  operation  of  obtaining  a  derivative  is  exceedingly 
laborious,  but  a  law  may  be  formulated  for  each  type  of 
primitive.  A  list  of  primitives  and  their  corresponding 
derivatives  is  given  in  Table  XXXV  and  may  be  referred 
to  for  use  or  verification.  Table  XXXV  contains  the 
differential  form  (40)  in  which  the  rate  of  the  dependent 
variable  is  expressed  in  terms  of  the  rate  of  the  independent 
variable. 

Derivative  Form  Differential  Form 

(39)         -^-  =  anxn-1.  (40)  dy  =  anx«~1dx. 

(40)  expresses  the  rate  of  change  of  y  in  terms  of  the 
rate  of  change  of  x.  In  other  words,  the  ordinate  changes 
anxn~l  times  as  fast  as  the  abscissa.  Suppose  a  =  3  and 
n  =  2,  then  (31)  becomes  (41),  (39)  becomes  (42),  and  (40) 
becomes  (42a): 

(41)     y  =  3x2,  (42)     ^  =  6x,  ^42a^     dy  =  Qxdx- 

The  following  are  obtained  from  (41),  (42),  and  (42a) : 

When  x  =  l,  then  the  ordinate  =  3, 
the  slope  =  6, 

the  rate  of  the  ordinate  =  6  times  the  rate  of 
the  abscissa. 
When  x  =  2,  then  the  ordinate  =  12, 
the  slope  =  12, 

the  rate  of  the  ordinate  =  12  times  the  rate  of 
the  abscissa. 
When  x  =  S,  the  ordinate  =  27, 
the  slope  =  18, 

the  rate  of  the  ordinate  =  18  times  the  rate  of 
the  abscissa. 

13.  A  primitive  curve  is  transformed  into  a  derived 
curve  which  we  call  the  derivative  by  the  operation  called 
differentiation.     The    separation   of   the  rates    transforms 


EATES,  DERIVATIVES,  AND  INTEGRALS         447 

the  derivative  into  the  differential  form.  The  reverse 
operation  which  transforms  the  differential  equation  into 
the  primitive  is  called  antidifferentiation  or  integration. 
Differentiation  is  therefore  the  process  of  analyzing  or  dis- 
integration of  an  algebraic  expression  into  its  tangents, 
whereas  integration  is  the  process  of  synthesizing  or  sum- 
ming and  building  a  curve  from  its  tangents.     The  symbol 

for  integration  is  I  which  was  originally  an  old-fashioned  S, 

suggested  by  the  summing  process.  It  also  resembles  the 
familiar  scroll  figures  on  musical  instruments.  Placing  the 
integral  symbol  in  (40)  we  obtain  (43)  which  is  an  instruc- 
tion to  perform  the  operation  of  antidifferentiation)  i.e., 
integration : 

(43)  fdy=fanxn-ldx. 


(43a)  Jdy=j4/  =  y. 


In  the  left-hand  member  of  (43)  y  is  subject  to  two 
opposite  operations  and  since  the  one  follows  immediately 
after  the  other  y  remains  unaffected.  The  two  operations 
are  in  juxtaposition  and  are  therefore  in  the  nature  of  a 
command  and  its  countercommand  and  therefore  they 
destroy  each  other  and  are  struck  out  as  is  shown  in  (43a) . 
An  indicated  operation  in  an  equation  can  be  removed  either 
by  performing  the  operation  or  subjecting  the  equation  to 
the  antioperation.  The  latter  method  was  applied  to  the 
left  member  of  (43)  and  the  former  method  was  applied  to 
the  right-hand  member  of  (43).  To  integrate  the  right- 
hand  side  of  (43)  reverse  the  law  of  differentiation  and  obtain 
the  following  law  for  the  integration  of  the  power  of  a  variable. 
Increase  the  exponent  by  unity  and  divide  the  coefficient 
by  the  new  exponent,  as  shown  in  (436) . 


nnxn~l+l 


(436)  y  =  j  anxn~ldx  =  - 

(31)  .*.     y  =  axn. 


448  PRACTICAL  MATHEMATICS 

Observation.  Every  equation  is  the  derivative  of  some 
primitive  equation.  The  former  may  be  obtained  from  the 
latter  by  graphic  methods  and  by  the  laws  of  differentiation. 

Every  equation  is  also  the  primitive  of  some  derivative 
equation.  The  primitive  is  also  called  the  integral  and  may 
be  obtained  by  graphic  methods  or  by  the  laws  of  anti-differen- 
tiation or  integration. 

14.  In  paragraph  8  it  was  shown  that  the  derivative  of  a 
sine  curve  is  represented  by  a  cosine  curve.  Therefore  the 
primitive  of  a  cosine  curve  is  a  sine  curve. 

Differentiation.  Integration. 

(44)  y  =  sin  x.  (46)  dy  =  cos  xdx. 

(45)  -~  =  cos  x  (47)       J  dy  =  I  cos  xdx. 

(46)  .*.     dy  =  cos  xdx.  (44)    .*.     y  =  sin  x. 

The  process  of  integration  is  simplified  by  recognizing 
the  primitive  forms  which  are  associated  with  definite 
derivatives  or  differential  forms  according  to  Table  XXXV. 
The  process  of  integration  is  performed  automatically  by 
various  instruments  known  as  water-meters,  gas-meters, 
integrating  wattmeters,  integraphs,  and  planimeters. 

Ex.  17.  Consult  Table  XXXV  and  write  the  derivatives  for 
the  following  primitives: 

(a)  i  =5  sin  6;        (6)  i  =  50  sin  6;        ,     (c)  i  =1  sin  50x2; 

(d)  i  =  .5  sin  t;       (e)  i  =  .5  sin  120x£;      (/)  y  =5  sin  (2x  +5). 

Ex.  18.  Consult  Table  XXXV  and  write  the  primitives  for  the 
following  derivatives: 

di  dz 

(a)     —  =5  cos  6;  (6)     —=-50  sin  6; 

d  i  di 

(c)     -=50x7  cos  50x«;  (d)    —=  -500x  cos  120x2; 


dt  '  v  '     dt 

!=60eoSK);  <*    S 


(/)    ^=60cos(W^);  (flf)    ^=5nsin(rc*+x). 


RATES,  DERIVATIVES,  AND  INTEGRALS         449 

Ex.  19.    Consult  Table  XXXV  and  write  the  primitives  for  the 

following  derivatives : 

(«)    g-5*«;  (6)    %=5T+K;  (c)    g-«r»+6r+e. 

(rf)  s—  ;  (e)  s— 1-25s     :  (/)  s— *■ 


15.  Logarithmic  and  Exponential  Formulas  are  inti- 
mately related,  as  shown  by  the  various  transformations 
which  follow,  and  also  by  consulting  Table  XXXV. 


(48) 

y  =  c*. 

(49) 

log.  i/  =  xlog. 

c. 

(50) 

log,  c  =  a  constant  = 

*K. 

(51) 

.*.     log,  y=xK. 

(52) 

y=?K. 

DlFFERENTATION 

Integration 

(52) 

»-w*. 

(54) 

dy=KzxKdx. 

(53) 

ax 

(55) 

/*-/*? 

(54) 

dy-K  zxKdx  =  Kydx. 

(52) 

KsxK 

y=    K     " 

(51) 

\oge  y  =  xK. 

(57) 

^=Kdx. 

y 

(56) 

P=Ky. 

dx       " 

(58) 

fdHKdx- 

(57) 

^  =  Kdx. 

(51)log«y=Xx. 

450 


PRACTICAL  MATHEMATICS 


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RATES,  DERIVATIVES,  AND  INTEGRALS         451 


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RATES,  DERIVATIVES,  AND  INTEGRALS         455 


16.  Graphic  Integration.  In  paragraph  8  it  was  shown 
by  means  of  Fig.  178  that  a  primitive  which  is  repre- 
sented by  a  sine  curve  has  a  derivative  which  is  a  cosine 
curve,  i.e.,  the  derivative  is  also  a  sine  curve  which  leads  the 
primitive  by  90°.  The  slope  of  the  primitive  at  any 
instant  of  time  is  determined  by  the  tangent  to  the 
curve,  at  the  point  whose  abscissa  corresponds  to  that 
definite  instant.  The  numeric  values  of  the  slopes  are 
plotted  as  the  simultaneous  ordinates  of  the  derivative 
curve.  The  reversal  of  graphic  differentiation  is  called 
graphic  antidifferentiation  or  graphic  integration  and 
implies  the  building  of  a  primitive  from  a  given  derivative. 


Fig.  180. — Graphic  Integration. 

In  Fig.  180  2/1  =  sin  0  is  a  given  derivative  curve  and 
ij2=—  cos  0  is  its  corresponding  integral.  The  numeric 
values  of  the  ordinates  of  2/1  are  equal  to  the  slopes  of  the 
tangents  to  y2.  The  ordinates  at  J',  K',  U,  M'  of  2/1  are 
equal  to  0,  .5,  .866,  and  1  respectively.  The  lines  0,  k,  I, 
m  at  the  left  of  the  Y  axis  have  slopes  equal  to  0,  .5,  .866, 
and  1  respectively.  At  the  points  J,  K,  L,  and  M  of  2/2 
the  tangents  are  respectively  parallel  to  0,  k,  I,  m  and  there- 
fore their  respective  slopes  are  0,  .5,  .866,  and  1.  The 
ordinates  of  P',  Q',  and  R'  on  2/1  are  .866,  .5,  and  0  respect- 
ively and  therefore  at  the  points  P,  Q,  and  R  on  2/2,  the 
tangents  are  parallel  to  I,  k,  and  0  respectively.  The 
integral  curve  is  therefore  a  sine  curve  which  touches  the 
series  of  tangents  at  the  points  J,  K,  L,  M,  P,  Q,  R. 


456  PRACTICAL  MATHEMATICS 

The  integral  curve  is  determined  as  follows:  The  small 
triangular  areas  which  have  been  added  outside  the  deriv- 
ative curve  are  constructed  so  as  to  be  equal  to  the  corre- 
sponding triangular  areas  subtracted  inside  the  derivative 
curve.  The  eye  can  adjust  these  pairs  of  triangular  areas 
very  accurately  after  a  little  practice  so  that  the  total 
error  is  negligible.  The  result  is  that  the  area  under  the 
derivative  curve  y\  is  replaced  by  the  area  under  the  stepped 
figure  which  is  bounded  by  risers  and  levels,  i.e.,  by  vertical 
and  horizontal  lines.  The  ordinates  of  the  stepped  figure 
change  only  at  the  levels  and  are  constant  between  the 
risers.  The  ordinates  of  the  levels  through  /',  K',  V , 
M',  P',  Q',  Rf  are  0,  .5,  .866,  1,  .866,  .5,  and  0  respectively, 
and  the  same  numeric  values  of  the  slopes  of  the  tangents 
of  2/2  are  observed  at  the  respective  points  J,  K,  L,  M, 
P,  Q,  and  R, 

The  construction  of  the  integral  curve  follows:  Extend 
the  risers  and  also  the  ordinates  passing  through  the  points 
J',  K',  L',  M',  P',  Qf,  and  R'.  Locate  a  pole,  i.e.,  a  fixed 
point,  at  a  unit's  distance  to  the  left  of  the  origin.  Extend 
the  levels  to  the  left  and  join  their  points  of  intersection  on 
the  Y  axis  with  the  pole  and  label  these  oblique  lines  o, 
k,  Z,  and  m.  Locate  J  at  a  unit's  distance  below  the  origin. 
Draw  a  tangent  parallel  to  o  extending  from  J  to  the  first 
riser  and  from  its  extremity  draw  a  tangent  parallel  to  k 
extending  to  the  second  riser.  From  the  last  point  draw 
a  tangent  parallel  to  I  extending  to  the  third  riser.  Con- 
tinue to  draw  tangents  between  consecutive  risers.  The 
steps  on  the  right  half  of  the  loop  were  constructed  to  have 
the  same  levels  as  the  steps  on  the  left  half  of  the  loop. 
The  tangents  for  the  loop  therefore  began  with  o  and  were 
followed  in  order  by  k,  I,  m,  I,  k,  and  o.  The  integral  curve 
is  then  drawn  so  as  to  touch  the  tangent  lines  at  J,  K, 
L,  M,  P,  Q,  and  R,  which  are  the  respective  points  of  inter- 
sections of  the  tangents  with  the  extended  ordinates  through 
J',  K',  L',  Mf,  P',  Q',  R'.    The  tangents  to  the  positive 


RATES,  DERIVATIVES,  AND  INTEGRALS         457 


loop  of  yi  have  positive  slopes,  whereas  the  tangents  to 
the  negative  loop  of  yi  have  negative  slopes.  The  tangents 
to  the  lower  loop  are  determined  by  constructing  parallels 
to  another  set  of  oblique  lines  drawn  from  the  pole  to  those 
points  on  the  Y  axis  which  are  crossed  by  the  lower  levels, 
as  shown  in  Figs.  181  and  182.  The  integral  of  a  sine 
curve  is  a  sine  curve  lagging  90°  behind  the  derivative, 

Method  of  Tangents 
a»=0  y=-lto  x-2   y-1 

The  ai-ea  is  represented  by  the  oi-dinates         q 
'      of  the  parabola  O  C  W  E  F 
P^poleH(O-l) 


Fig.  181. — The  Graphic  Integration  of  a  Linear  Graph. 

as  expressed  in  (59)  and  (60),  and  which  is  also  substan 
tiated  by  form  14  in  Table  XXXV. 

Derivative  Integral 

dy2 


(59) 


yi  =  sm  t»  = 


d8* 


(60)  2/2  =  —  cos  6  =  sin  I  6  —  rr 


17.   The    Graphic    Integration    of   a   Linear    Equation. 

In  Fig.  181  the  linear  graph  K'G'  is  represented  by  Eq.  (61) 
and  its  integral  by  the  parabola  OCWEF,  whose  equation 
is  expressed  by  (62): 

Derivative  Integral 

(61)         yi=-l+x.  (62)        y2=~x+ix2. 


458  PRACTICAL  MATHEMATICS 

The  area  of  the  stepped  figure  K'KH"HA"A  replaces 
the  negative  area  between  the  X  axis  and  the  line  K'A, 
whereas  the  stepped  figure  AA'BB"GG'  replaces  the  positive 
area  between  the  X  axis  and  the  line  AG' '.  The  levels  are 
projected  over  to  the  Y  axis  and  the  intersections  of  the 
former  on  the  latter  are  joined  with  the  pole  P,  which  is 
located  at  a  unit's  distance  to  the  left  of  the  origin.  The 
tangent  OR  is  parallel  to  k  and  extends  from  the  origin 
to  the  first  riser  TK.  The  tangent  RH  is  parallel  to  h  and 
extends  between  the  first  and  second  risers.  The  tangent 
HA"  is  parallel  to  a  and  extends  between  the  second  and 
third  risers.  The  tangent  A"Q  is  parallel  to  b  and  extends 
between  the  third  and  fourth  risers.  The  tangent  QF  is 
parallel  to  g  and  extends  from  the  fourth  riser  to  F,  which 
is  the  last  point  on  the  parabola.  The  parabola,  i.e.,  the 
integral  curve,  touches  the  tangents  at  0,  C,  W,  E,  and  F, 
which  are  obtained  by  projecting  K' ',  H',  A,  B',  G'  from  the 
derivative  curve.  The  ordinates  of  the  integral  curve 
measure  the  area  bounded  by  the  line  K'G' ,  the  X  axis, 
the  Y  axis,  and  the  extended  ordinate.  Thus  VC  measures 
the  area  K'H'VO.     VC=  -.375  and  the  area  of 

K'H'VO=-^f-(OK'+VH')  =  -^(l  +  .5)  =  -.375. 

-i  Li 

The  ordinates  VC,  OK',  and  VH,  as  well  as  the  area  of 
K'H'VO  are  negative,  since  they  extend  below  the  X  axis. 
The  ordinate 

aut          e             nr>A      -OK'XOA          1X1 
AW  =  —  .5  =  area  OK' A  = = = o- • 

The  ordinate  ME  =  —  .375  =  the  excess  of  the  negative 
area  OK' A  over  the  positive  area  AMB'' 

Oi£'A-AM£'=-.5+(.5X.5X.5)  =  -.375. 

The  ordinate   at  F  equals  zero,  which  indicates  that  the 
excess  of  the  negative  area  OK' A  over  the  positive  area 
AFG'  is  zero. 


RATES,  DERIVATIVES,  AND  INTEGRALS         459 

18.  Double  Integration.  Any  curve  may  be  regarded 
as  either  a  derivative  or  a  primitive.  Thus  the  parabola 
CHEI'Z'G  in  Fig.  182,  is  the  integral  of  the  linear  graph 
B'Al.     The  cubic    curve    VHJKLMN  is  the  integral  of 


C  D  E  F  G  is  the  integral  of  A  B 

i   CDEFG 


I'    Mtf 


Fig.  182. — Double  Integration. 

the  parabola  CHEI'Z'G.  In  this  sense  the  parabola  is 
the  first  derivative  of  the  cubic  and  the  linear  graph  is  a 
second  derivative  of  the  cubic,  since  it  is  a  derivative  of  a 

derivative.     The  symbol  for  the  second  derivative  is  -r-  -r- 
J  ax  ax 


460  PRACTICAL  MATHEMATICS 


d2 
which  contracts  into  -^  but  is  not  the  same  as  ( -£- )  ,  since 


(A)2 

\dx)  ' 
the  latter  is  obtained  by  squaring  the  first  derivative. 


Linear  Graph  A  Derivative 

(63)  Vl  =  -l+x,  (63a)        ^-=-l+x, 

A  Second  Derivative 
(636)  jp(v3)  =  -l+*. 

The  Parabola  A  Derivative 

(64)  y2  =  .25-x+.5x2.  (64a)    j-(y3)  =  .25-x+.5x2. 

The  Cubic 

(65)  y3  =  .25x-.5x2+ixs. 

The  symbol  of  double  integration  is  I    J  ,  as  shown  in 

(66).      If  one  integration  is  performed,  as  shown  in  (67), 
then  double  integration  reduces  to  single  integration    and 

(66)  becomes  (68)  and  the  latter  reduces  to  the  cubic  (65). 
The    innermost    integral    symbol    operates    upon    the 

expression  as  far  as  and  including  the  innermost  differ- 
ential, as  shown  by  the  brace  in  (66). 


(66)      y3=fj(-l+x)dxdx. 


(67)  J  (  —  \-\-x)dx  =  —  J  dx-\-  J  xdx+a,  constant 

=  .25-x+.5x2. 

The  derivative  of  a  constant  is  zero.  Therefore  an 
arbitrary  constant  may  be  added  to  the  integral  after 
performing  an  integration.  The  value  of  the  constant 
corresponds  to  the  Y  intercept  of  the  integral  graph.  The 
parabola  for  convenience  and  clarity  was  constructed 
from  the  point  C  on  the  Y  axis  and  therefore  the  constant 


RATES,  DERIVATIVES,  AND  INTEGRALS         461 

of  integration  =  .25,   which   is  the  numeric    value  of  the 
Y  intercept. 

(68)     2/3=  f(.25-x+.5x2)dx  =  f.25dx-  (xdx+  C.dx^dx. 

(65)      y3  =  .25x-.5x2+^. 

Observation.  The  degree  of  the  integral  is  one  greater 
in  unit  value  than  the  degree  of  its  derivative  when  both  repre- 
sent algebraic  equations.  When  an  integral  curve  passes 
through  the  origin  then  its  ordinate  measures  that  area  under 
its  derivative  curve,  which  is  bounded  by  the  simultaneous 
ordinate  of  the  derivative,  the  derivative  curve,  and  the  two 
axes. 

When  the  integral  curve  does  not  pass  through  the  origin 
then  the  area  of  the  derivative  is  measured  by  subtracting  the 
ordinate  of  the  integral  curve  from  the  numeric  value  of  the 
latter's  Y  intercept.  In  other  words,  a  horizontal  line  is 
drawn  through  the  Y  intercept  and  the  distance  between  it 
and  the  integral  curve  measures  the  area  of  the  derivative 
curve  at  a  simultaneous  ordinate. 

19.  The  Scales  of  Integral  Curves.  If  a  derivative  curve 
is  plotted  so  that  its  X  and  Y  scales  are  equal  then  the  same 
scales  apply  to  the  integral  curve.  This  is  shown  in  Figs. 
180  and  181.  In  Fig.  182  the  vertical  scale  of  the  linear 
graph  is  one-half  of  its  horizontal  scale.  In  such  cases 
when  the  unit  polar  distance  X'O  is  constructed  according 
to  the  horizontal  scale,  then  the  scales  of  the  integral  curve 
(parabola)  will  be  like  the  scales  of  its  derivative.  The 
polar  distance  00'  for  integrating  the  parabola  in  Fig.  182 
is  one-fourth  of  the  horizontal  scale,  and  therefore  the 
vertical  scale  of  the  cubic  is  reduced  to  one-fourth  of  the 
vertical  scale  of  the  parabola.  Therefore  the  vertical  scale 
of  the  cubic  equals  one-eighth  of  its  horizontal  scale.  The 
ratio  of  the  polar  distance  to  a  unit  distance  on  the  hori- 
zontal axis  is  called  the  polar  scale  factor. 


462  PEACTICAL  MATHEMATICS 

Observation.  In  the  process  of  graphic  integration  the 
horizontal  scales  of  both  integral  and  derivative  curves  are 
identical.  The  vertical  scales  of  the  integral  and  derivative 
curves  are  in  the  same  ratio  as  the  polar  scale  factor. 

Ex.  20.  Integrate  the  sine  curve  yx  =sin  6  as  shown  in  Fig.  180, 
wherein  y2  is  the  integral  of  yu  Construct  a  straight  line  JZ 
through  U'  from  which  measure  the  height  of  the  points  of  y2. 
The  latter  give  the  areas  between  the  sine  curve  y\  and  the  x  axis 
from  the  origin  to  the  simultaneous  ordinate.    Thus  L  is  .5  of  a 

X 

unit  above  the  line  JZ,  and  therefore  the  area  of  J'L'~  =  .5  square 

o 

units.    In  other  words,  the  ordinate  of  the  sine  curve  has  swept 

through  or  generated  an  area  of  |  square  unit.     Determine   (a) 

the  area  of  half  a  sine  loop;   (6)  the  area  of  a  loop;   (c)  divide  the 

area  of  a  loop  by  x  its  length  and  determine  the  mean  value  of  its 

ordinates;   (d)  the  area  for  one  cycle.    Why  do  the  ordinates  of 

2/2  decrease  after  passing  R. 

Ex.  21.  In  Fig.  182  draw  a  parallel  CG  to  XX'  through  C. 
Measure  the  segments  of  five  perpendiculars  extending  between 
the  parabola  and  CG  and  show  that  their  numeric  values  express 
the  area  between  B'B  and  XX'  and  from  the  measured  perpendicular 
to  the  origin. 

Ex.  22.    Integrate  the  following  graphs: 

(a)  y=.5x;         (b)  y  =  .75x;        (c)  y=x;        (d)  y  =  1.25x; 

(e)  y  =  1.5x;      (/)  y=2x;  (g)  y=Sx;      Qi)  y=4x; 

(i)   y=5x;         (j)  y=6x;  (k)y=7x;      (I)   y  =  10x. 

In  each  case  begin  the  integral  curve  at  the  origin,  but  verify  the 
work  by  checking  the  ordinate  of  the  integral  with  the  area  of 
the  triangle. 

Ex.  23.    Integrate  the  following  graphs: 

(a)y=-l+x;  (b)  y  =  -.7o+x;  (c)  y=  -.5+x; 

(d)  y=-.25+x;  (e)  y=.25+x;  (/)  y  =  .5+x; 

(g)  y  =  .75+x;  (h)  y=-l+2x;  (i)  y=-l+.5x; 

U)  V  =  -5 -x)  (k)  y  =  l-2x;  (l)  y=-.l-.5x; 

(m)  y=  —,5  —  .lx. 


RATES,  DERIVATIVES,  AND  INTEGRALS         463 


In  each  case  draw  two  integral  curves  so  that  one  integral  inter- 
cepts the  Y  axis  at  the  origin  and  so  that  the  other  intercepts 
the  Y  axis  .25  above  the  origin.  Use  a  polar  factor  of  .5  for  the 
second  of  the  two  integral  curves.  Write  the  equations  for  the 
integral  curves. 

Ex.  24.  Perform  graphic  differentiation  upon  the  two  integral 
curves  in  Ex.  23,  showing  that  both  integral  curves  give  the  same 
derivative. 

Ex.  25.    Integrate  the  integral  curves  obtained  in  Ex.  22. 

Ex.  26.    Integrate  the  integral  curves  obtained  in  Ex.  23. 

Ex.  27.  Construct  a  sine  curve  y  =sin  8  as  shown  in  Fig.  183, 
and  measure  its  ordinates  at  the  points  0,  A",  B",  C",  D",  E", 
F",    G",    H",   etc.     Square   the   respective   ordinates.     Plot   the 


Fig.  183.— The  Integral  of  F=sin2  6. 

respective  squares  on  the  simultaneous  ordinates  obtaining  the 
points  A'B'C'D"E'F'G'H",  etc.,  which  lie  upon  the  sin2  curve 
whose  equation  is  expressed  in  (68). 


(68) 


?/=sin2  6  =  (sin  9); 


The  graph  of  (68)  will  be  readily  identified  with  the  power 
curves,  shown  in  Figs.  169  and  170.  Integrate  the  graph  of  (68), 
which,  is  shown  as  ABCDEFGH,  etc.,  in  Fig.  183.  Determine 
the  area  of  the  sin2  6  curve  under  one  loop  and  divide  this  value 
by  x  the  length  of  the  loop.  The  quotient  is  the  average  height 
of  the  sine  curve,  and  the  square  root  of  the  latter  is  defined  as 
the  mean  effective  value  of  the  sine  curve.  The  graph  of  (68)  is 
also  a  sine  curve  whose  axis  lies  midway  between  D  and  the  axis 
of  the  sine  curve  and  therefore  the  area  of  (68)  may  be  obtained 
from  (a)  and  (6)  in  Ex.  20. 


464  PRACTICAL  MATHEMATICS 

20.  The  Integration  Formula  for  the  Area  of  a  Curve. 

In  the  preceding  paragraphs  we  have  observed  that  the 
successive  ordinates  of  the  derivative  curve  represent  the 
successive  rates  of  change  of  the  integral  curve  and  inversely 
the  successive  ordinates  of  the  integral  curve  represent  the 
successive  areas  under  the  derivative  curve.  In  all  such 
cases  if  we  designate  the  ordinates  of  the  integral  and 
derivative  curves  by  y\  and  y  respectively,  we  have  the 
relations  expressed  in  (69),  (70),  and  (71). 

(69)  y  =  ^  (70).  dVl=ydx. 


(71)  yi=Jydx. 


(71)  is  called  the  areal  formula,  i.e.,  the  integration 
formula  for  obtaining  the  area  of  any  curve  whose  ordinate 
is  represented  by  y.  The  area  (74)  under  the  curve  ODB 
in  Fig.  184,  whose  equation  is  expressed  in  (72),  is  obtained 
by  integrating  (73)  after  substituting  the  value  of  y  in  the 
areal  formula  (71). 

(72)  yi=axn. 

(73)  2/1  =  area  under  ODB  =  I  axndx. 

axn+l 


(74)  yi  = 


n+1 


(73)  may  be  interpreted  as  the  limit  of  the  sum  of  all 
the  elementary  strips,  such  as  D  in  Fig.  184,  whose  dimen- 
sions are  y  and  dx.  An  elementary  strip  represents  the 
rate  of  change  of  the  area  of  the  derivative  curve.  These 
strips  extend  from  the  origin  indefinitely  to  the  right  and 
therefore  (73)  expresses  an  indefinite  area  and  accordingly 
formulas  (71)  and  (73)  are  called  indefinite  integrals. 


RATES,  DERIVATIVES,  AND  INTEGRALS 


465 


If  it  is  desired  to  limit  the  area  between  the  origin  and 
the  ordinate  whose  abscissa  is  c,  this  fact  defines  an  area 
and  prescribes  a  corresponding  definite  integral.  The 
limits  are  written  adjacent  to  the  integral  symbol,  as  shown 
in  (75). 


(75) 


yi=JoydX  =  ^\o 


Fig.  184. — The  Areal  Integral  Formula. 

The  right-hand  member  of  (75)  is  evaluated,  i.e.,  its 
value  is  determined  by  substituting  the  limiting  value  c 
in  place  of  x,  and  therefore  (75)  becomes  (76) : 


(76) 


2/i  =  area  (from  0  to  c) 


ac 


n+l 


w+1 


For  the  special  equation  y  =  2x3  the  area  between  the 
origin  and  an  ordinate  whose  distance  =  1.5  units  is  expressed 
in  (77): 

2(1.5)4 
4 


(77) 


2/1 


2.53+square  units. 


466  PRACTICAL  MATHEMATICS 

If  the  area  extends  from  the  origin  to  an  ordinate  b 
whose  abscissa  =  4  units,  these  facts  are  symbolized  in  (78), 
which  evaluates  to  (79): 

r  *  axn+iv  ahn+i 

(78)  yi=  I    ydx  =  - 


ri  +  1  J0     n+1 


(79)  2/1  =  -M-  =  128  square  units. 


The  area  between  the  two  ordinates  whose  respective 
abscissas  are  b  and  c  is  indicated  in  (80) : 

(80)  yi_  j  ,*-§j]; 

The  evaluation  of  (80)  gives  us  the  difference  between 
the  values  obtained  in  (76)  and  (78).  Therefore  (80) 
becomes  (81),  which  reduces  to  (82): 

axn+lV     a(b)n+1     a(c)n+1 

(81)       2/i=^+iJc=^+r-~n+r- 

(82)  2/1  =  128-2.53  =  125.47  square  units. 

Observation.  A  definite  integral  is  a  prescribed  summation 
and  is  pictured  by  a  definite  area  under  a  curve.  The  limits 
of  the  integral  indicate  the  extent  of  the  summation  process. 
The  integrand  or  quantity  affected  by  the  integral  sign  is  the 
product  of  the  ordinate  of  a  curve  times  the  rate  of  change  of 
its  abscissa.  The  limiting  values  of  the  abscissas  of  the  curve 
are  placed  to  the  right  of  the  integral  symbol,  so  that  the  greater 
or  superior  limit  appears  above  and  the  lesser  or  inferior  limit 
appears  below  the  integrand.  The  integration  is  performed 
by  substituting  the  equivalent  form  from  the  table  of  integrals. 
The  evaluation  of  the  form  is  the  numeric  excess  of  the  two 
results  which  are  obtained  by  substituting  first  the  superior 
limit  and  secondly,  the  inferior  limit. 


RATES,  DERIVATIVES,  AND  INTEGRALS         467 

21.  The  area  of  the  sine  loop  (84),  as  shown  in  Fig.  185, 
may  be  considered  as  the  integral,  i.e.,  limit  of  the  sum  of 
the  elementary  strips  of  dimension  y  and  d0,  as  expressed 
in  (83).  The  limits  of  0  are  zero  and  x,  which  indicate  the 
extent  of  the  sine  loop  along  the  horizontal  or  8  axis. 

(83)  2/1  =  area  sine  loop  =  I    yd%. 

Jo 

(84)  2/  =  sinG. 


(85) 


2/i=  I    sin  0d0  =  —cos  0    . 

Jo  Jo 


(86)     /.     2/i  =  (-cosx)-(-cos0)  =  -(-l)-(-l) 

=  2  square  units. 


Fig.  185. — The  Area  of  the  Sine  Loop. 

The  average  value  (y&ver)  of  the  ordinates  of  the  sine 
loop  whose  equation  is  given  in  (84)  is  obtained  by  dividing 
the  area  of  the  sine  loop  by  its  length  as  expressed  in  (87) : 

fsin  0d0 

(87)  yaver=^=^ =  -  =  .636. 

%  xz 

The  maximum  ordinate  of  the  current  curve,  i.e.,  sine 
loop  in  Fig.  186  is  J,  and  therefore  its  average  value  =  .6367. 
In  other  words,  the  average  value  of  a  current  which  follows 
the  sine  law  is  .636  times  its  maximum  value. 

Ex.  28.  Determine  the  areas  under  the  following  curves, 
between  the  origin  and  the  abscissa  5,  between  the  origin  and  the 
abscissa  10,  between  the  origin  and    the  abscissa  1,  between  the 


468 


PRACTICAL  MATHEMATICS 


abscissas  5  and  10:  (a),  (b),  (c),  (d),  (e),  (/),  (g),  (A),  (i),  (j), 
in  Ex.  22;  (a),  (6),  (c),  (d),  (e),  (/),  (<?),  (h),  (i),  (j),  (*),  (Q,  (m), 
in  Ex.  23. 

Ex.  29.  Determine  the  area,  under  the  following  curves, 
between  the  origin  and  abscissa  1,  between  the  origin  and  abscissa 
2,  between  the  abscissas  1  and  2:  (a),  (b),  (c),  (d),  (e),  (/),  in 
Ex.  7;   (a),  (6),  (c),  (d),  (e),  (/)  in  Ex.  8;   (d),  (e),  (/),  in  Ex.  19. 

Ex.  30.  Determine  the  area  under  one  loop  of  the  following 
curves:  (a),  (b),  (c),  (d),  in  Ex.  10;  (a),  (6),  (c),  (d),  (e),  (/), 
in  Ex.  17;   (a),  (6),  (c),  (d),  (e),  (/),  (tti  in  Ex.  18. 

22.  Mean  Effective  Value  of  a  Harmonic  E.M.F.  or 
Current.  The  heating  effect  representing  the  useful  energy 
spent  in  a  circuit  varies  as  the  square  of  the  current.  In 
a  direct-current  circuit  we  recognize  the  familiar  expression 


Elementary  Strip 


Effective  Valueleff.         Avtnat  Value  IAve. 


Fig.  186. — The  Distinction  between  Average  and  Effective  Values. 


PR  loss.  An  equivalent  loss  may  take  place  with  an 
alternating  current.  There  is  this  distinction,  however, 
that  with  the  direct  current,  J  remains  constant  and  there- 
fore I2  remains  constant,  whereas  with  alternating  current  I 
is  variable  and  therefore  I2  is  different  at  each  instant.  In 
order  to  produce  the  same  energy  effect  the  average  of  the 
(72alt)  of  the  alternating  current  must  equal  the  (72dir)  of 
the  direct  current.  The  square  root  of  the  average  of  the 
squares  of  the  successive  instantaneous  values  of  the  alter- 
nating current  is  called  its  effective  value  (7eff ) . 


(88) 


(89)   Jeff 


/2dir  =  /2, 


■4 


sum  of  squares  of  ordinates        /  area  sin2  curve 


number  of  ordinates 


J 


length  sin2  curve' 


RATES,  DERIVATIVES,  AND  INTEGRALS         469 

The  average  of  the  squares  may  be  obtained  by  inte- 
grating the  area  under  the  curve  of  squares  of  current 
and  dividing  the  area  by  the  length  of  the  loop. 

(90)  is  the  equation  of  an  alternating  current  and  (91) 
the  corresponding  equation  of  its  square.  These  curves 
are  illustrated  in  Fig.  183.  The  area  under  the  sin2  curve 
is  given  in  (92) : 

(90)  i  =  J»sin8. 

(91)  ;2  =  /TO2sin20. 


(92) 


Area=  j  *  Im2  sin2  0d0  =  Im2  J  'sin2  6d8. 

fljT       A         (Im2  »  Im2  ft\ 

=  v  2 — ) ~ \T cos    — r cos  / 


J    2T        7    2        7    2        7    2T 

"IT  'IT-1-  4  =     2  ' 


(93)  Jeff 


(94)  results  from  substituting  (92)  and  (93).  The  inter- 
pretation of  (94)  states  that  the  effective  value  of  an  alter- 
nating current  equals  .707  times  its  maximum  value.  The 
effective  value  Ees  of  an  E.M.F.  is  the  square  root  of  the 
average  of  the  squares  of  the  successive  instantaneous 
values  of  the  alternating  E.M.F.  It  may  be  derived  in 
the  manner  described  for  obtaining  Ies  or  by  considering 


470  PRACTICAL  MATHEMATICS 

EeS2 

the  energy  loss  equal  to  -=-,  as  shown  in  (95),  (96),  and 
(97): 

(95)  I*?R=%f. 

(96)  Ies2R2  =  Ee^. 

(97)  EeS  =  IesR. 

The  average  and  effective  values  of  a  sine  wave  are 
shown  in  Fig.  186. 

23.  The  ordinates  of  the  sine  curve  are  represented  in 
Fig.  187  by  the  two  sets  of  perpendiculars  P\R,  PR,  P2R2, 
P3R3,  etc.,  and  QiSh  QS,  Q2S2,  Q3S3,  etc.  If  the  total 
number  of  ordinates  is  designated  by  n,  then  the  ordinates 

71 

may  be  arranged  in  —  pairs  of  ordinates,  such  as  PR  and 

QS,  whose  corresponding  radii  vectors  are  90°  apart  and 
therefore  PR  =  I  sin  0  and  QS  =  I  cos  Q.  The  sum  of  the 
squares  of  each  pair  of  ordinates  multiplied  by  the  number 
of  pairs  of  ordinates,  is  the  sum  2  of  the  squares  of  all  the 
ordinates  as  expressed  in  (98).  2  is  the  symbol  of  summa- 
tion of  a  number  of  quantities  which  are  similarly  constituted. 
In  this  case  each  pair  of  ordinates  is  of  the  type  PR  and  QS. 

in     .       ,                ,      ,.     ,         2PR2+QS2 
Sum  of  «  pairs  of  squares  of  ordinates      5 

total  number  of  ordinates  n 

_  ZZOT2(sin2  6+cos2  6)  =2Tm2  =  nlm2  =  Im2 
2n  2n  "  2n       2  ' 

(99)  7eff  =  ^  =  .707/m. 

24.  The  Moment  of  Inertia  Formula  for  a  Structural 
Section.  The  moment  of  inertia  of  a  plane  figure  is  expressed 
in  (100),  which  is  the  integral  of  the  products  obtained  by 
multiplying  each  elementary  area  by  the  square  of  its 
distance  from  the  neutral  axis. 


RATES,  DERIVATIVES,  AND  INTEGRALS         471 

In  Fig.  188  the  rectangular  section  has  a  width  b  and 
a  height  l-^J  above  the  neutral  axis  and  a  negative  height 

(  —  7T  J  below  the  neutral  axis. 

(100)  1=1  distance  X  elementary  area=  jifxdy. 


Y b > 

2 

cly 

J 

Elementary  Strip  * 

t 

y 

\ 

J 

) 

Fig.  187. — The  Geometric  Representa-  Fig.  188. — The  Determination 
tion  of  the  Effective  Value  of  a  Sine  of  the  Moment  of  Inertia  of 
Loop.  a  Rectangle. 

The  elementary  strip  in  Fig.  188  has  dimensions  b  and 
dy  and  therefore  an  area  bdy.  The  distance  of  the  strip 
from  the  axis  is  y.      The  strips  extend  from  the  superior 

limit  ( -=-  J  to  the  inferior  limit  (  —  -~- ) .   Therefore  the  moment 

of  inertia  Jrect  of   a  rectangular  section   of   width   b  and 
depth  D  about  its  neutral  axis  is  expressed  in  (101) : 


(101) 


rect 


=  J    Dy2bdy  =  b  J    ^y2dy 
=  SJ_d     3V2/       3\     2/ 


472  PRACTICAL  MATHEMATICS 

(102)  irect~"2r+"24"~"l2"* 

25.  The  Static  Moment  Formula  for  Any  Figure.    The 

static  moment  of  a  plane  figure  is  expressed  in  (103),  which 
is  the  integral  of  the  products  obtained  by  multiplying 
each  elementary  area  by  its  distance  from  an  axis  of  refer- 
ence called  a  moment  axis: 

(103)  M=  J  distance  X  elementary  area  =  j  yxdy. 

For  any  plane  figure  the  distance  z  of  the  neutral  axis 
from  the  moment  axis  may  be  computed  by  (104),  which 
is  the  ratio  of  the  static  moment  of  the  figure  to  its  area. 

j  y%dy  _  static  moment  of  figure 
fydx  area  °f  nSure 

26.  Graphic  Determination  of  Static  Moments,  Moments 
of  Inertia,  and  Centers  of  Gravity  of  Sections.  In  Fig.  189, 
2/i  is  the  graph  of  a  sine  curve  (105)  and  A\{y)  is  its  integral 
(106).  Therefore  the  ordinates  of  A\{y)  represent  the 
areas  under  y\  between  the  origin  and  the  simultaneous 
ordinate  of  Ai(y). 

The  static  moment  of  a  sine  curve  about  the  Y  axis  is 
expressed  in  (107),  which  'is  obtained  from  (103)  by  sub- 
stituting 0  for  the  distance  of  the  elementary  strip  from 
the  Y  axis  and  yidQ  for  its  area: 

(105)    2/i  =  sin6.  (106)    Ai(y)  =  Jsin  0d0. 

(107)  M  =  ftiyidQ. 

(108)  M  =  JmdQ  =  A2(m). 

In  (107)  the  product  0i/i  is  replaced  by  m,  as  expressed 
in  (108).     If  each  ordinate  of  y\  is  multiplied  by  its  distance 


RATES,  DERIVATIVES,  AND  INTEGRALS         473 


0  and  the  respective  products  plotted  on  the  measured 
ordinates  we  shall  obtain  the  curve  m,  which  is  called  the 
moment  curve.  (108)  indicates  that  the  area  or  integral  of 
the  m  curve  will  give  the  static  moment  of  the  sine  curve. 
The  integral  of  the  m  curve  is  represented  by  A2(m),  and 
therefore  any  measured  ordinate  of  A2(m)  is  the  numeric 
value  of  the  moment  of  that  much  of  the  area  of  the  sine 


Sine  Curve  J/L 

Moment  Curve   Itl 

Area  Curve  A^V) 

-Area  Curve  Ao^) 


Fig.   189. — Graphic   Determination    of    Static    Moment,   Moment  of 
Inertia,  and  Center  of  Gravity. 

curve  which  extends  from  the  origin  to  the   simultaneous 
ordinate. 

The  distance  of  the  center  of  gravity  from  the  Y  axis 
for  any  area  of  the  sine  curve  extending  from  the  origin 
to  an  ordinate  is  equal  to  the  ratio  of  the  simultaneous 
ordinates  of  A2(m)  and  A i(y): 

/*/^vx    ^    -,        <•         .,  simultaneous  ordinate  A  2  (m) 

(109)    Center  of  gravity  y\  =  — tt r- — i — A  /  \« 

v       '  &         J  *       simultaneous  ordinate  A i{y) 

The  center  of  gravity  may  be  determined  by  producing 
the   last    tangent   of  the    second    integral   curve   until   it 


474  PEACTICAL  MATHEMATICS 

intersects  the  horizontal  axis.  The  center  of  gravity  will 
lie  on  the  ordinate  which  passes  through  the  point  of  inter- 
section on  the  X  axis.  The  static  moment  area  curve 
should  be  obtained  and  checked  by  integrating  the  given 
curve  twice  in  succession. 

The  moment  of  inertia  of  any  figure  may  be  obtained 
by  integrating  the  given  section  three  times  in  succession. 
Twice  the  ordinate  of  the  third  integral  curve  is  the  numeric 
value  of  the  moment  of  inertia  of  the  given  section.  We 
may  express  the  moment  of  inertia  of  any  plane  section 
about  the  Y  axis  by  (110)  and  the  moment  of  inertia  about 
the  X  axis  by  (111): 

(110)  Iy  =  jx2ydx.  (Ill)        Ix  =  jy2xdy. 

Let  y  be  the  last  ordinate  of  tne  given  section,  then  the 
simultaneous  ordinate  y\  of  the  integral  curve  measures 
the  area  of  y: 

(112)  yi=jydx. 

The  area  of  the  y\  curve  is  expressed  by  the  integral 
curve  (113),  whose  ordinate  is  yi\ 

(113)  2/2=  j  yidx=  I  I  ydxdx=  j  yxdx. 

The  area  of  the  2/2  curve  is  expressed  by  the  integral 
curve  whose  ordinate  is  yz,  as  expressed  in  (114)  and  (115) : 

(114)  ys==  I  y2dx=  J   J    lydxdxdx. 

(115)  ys  =  ffyxdxdx  =  J^j-  dx. 

The  three  successive  integrations  are  indicated  by  the 
triple  integral  sign  in  (114).  The  ordinate  1/3  of  the  triple- 
integration  curve  is  half  of  the  ordinate  of  the  curve  which 


RATES,  DERIVATIVES,  AND  INTEGRALS         475 

would  be  obtained  by  plotting  (110).     The  ordinate  of  \j2 
in  (113)  may  be  compared  with  (103). 

Observation.  If  a  given  curve  be  integrated  three  times  in 
succession  then  the  simultaneous  ordinates  of  the  four  curves 
determine  the  following  properties:  The  ordinate  of  the  first 
integral  curve  measures  the  area  of  the  given  curve,  the  ordinate 
of  the  second  integral  curve  measures  the  static  moment  of  the 
area,  and  the  ordinate  of  the  third  integral  curve  measures 
one-half  of  the  moment  of  inertia  of  the  area.  The  area  is 
measured  from  the  origin  to  the  simultaneous  ordinate  and  the  Y 
axis  is  the  axis  of  reference.  When  the  X  axis  is  used  as 
the  axis  of  reference  then  the  corresponding  pole  is  located  on  the 
Y  axis  at  a  unit's  distance  below  the  origin. 

Ex.  31.  Draw  the  half  section  of  a  T-rail  and  by  graphic 
integration  determine  its  area,  the  static  moment,  center  of  gravity, 
and  moment  of  inertia  from  the  neutral  axis. 

Ex.  32.  Draw  the  load  diagram  L  of  a  beam  by  constructing 
ordinates  above  the  X  axis  to  represent  the  loads  and  then  join 
the  ends  of  the  ordinates.  Integrate  L  which  gives  a  curve  S 
whose  simultaneous  ordinates  express  the  shear  on  the  beam 
at  that  section.  Integrate  S  obtaining  the  curve  B.  Any  simul- 
taneous ordinate  of  B  expresses  the  bending  moment  of  the  beam 
at  that  section.  Join  the  last  point  J  of  B  with  the  origin  0. 
Through  the  pole  draw  a  line  parallel  to  OJ  intersecting  the  Y  axis 
at  H.  Then  HO  is  the  force  acting  at  the  left  support.  How  can 
the  force  at  the  other  support  be  determined  from  the  diagram? 

Observation.  At  maximum  and  minimum  points  on  a 
primitive  curve,  the  tangents  are  horizontal  and  therefore  at  the 
corresponding  points  on  the  derivative  curve  the  ordinates  are 
zero.  Likewise  the  integral  curve  will  have  its  maximum  and 
minimum  points  on  the  simultaneous  ordinates  which  pass 
through  the  zero  points  of  a  derivative. 

27.  Nomenclature  and  Interpretation  of  Differential  and 
Integral  Expressions.  Velocity  is  the  time  rate  of  change 
of  linear  or  angular  space.  Therefore  linear  velocity  is 
denned  by  (116)  and  angular  velocity  by  (117): 


ds 

Data. 

2  =  time. 

v=Tf 

s  =  linear  space  traversed. 

<28 

0  =  angular  space  traversed. 
v  =  linear  velocity, 
a)  =  angular  velocity. 

476  PRACTICAL  MATHEMATICS 

(116) 
(117) 


Acceleration  (a)  is  the  time  rate  of  change  of  velocity, 
as  expressed  in  (118): 

Substituting  the  equivalent  for  v  from  (116)  we  obtain 
(119): 

/ ., .,  ftc  d  ds    d2s 

(119)  a=dtd=df- 

The  successive  derivative  symbols  are  not  multiplied 

d2 
but   are   abbreviated  as   a   second   derivative   -^    which 

means  the  operation  of  differentiation  is  to  be  performed 
twice  in  succession.  (119)  states  that  acceleration  is  the 
rate  of  change  of  space  per  second  per  second. 

Ex.  33.    A  linear  velocity  is  given    as    45    miles    per   hour. 

ds 
What  is  the  value  of  —  when  expressed  (a)    in  feet  per  minute 

(b)  in  feet  per  second? 

ds 
Ex.  34.    The  value  of  —  is  given  as  100  kilometers  per  hour. 

What  is  the  equivalent  velocity  expressed  (a)  in  meters  per  second; 

(b)  feet  per  second? 

dV 
Ex.  35.     What  is  the  meaning  of  — ,  wherein  V  represents 

Cut 

ds  dV 

volume?    Write  the  value  of  —  for  a  6-in.  round  pipe,  when  —  = 

CLZ  CLZi 

30  gallons  of  water  per  second. 

Ex.  36.    Write  the  symbolic  or  derivative  form  for  the  follow- 
ing and  show  their  values  by  differentiating  the  given  primitives: 


RATES,  DERIVATIVES,  AND  INTEGRALS         477 

(a)  Temperature  coefficient  is  a  rate  of  change  of  resistance  (R) 
with  respect  to  temperature  (T).     The  primitive  is 

R=Ro(l+M42T). 

(6)  The  coefficient  of  linear  expansion  is  a  rate  of  change 
of  length  (L)  with  respect  to  temperature  (T).  The  primitive  for 
hard-drawn  copper  wire  is 

L=Lo(l+.000009477). 

(c)  The  coefficient  of  expansion  is  the  rate  of  change  of  volume 
(V)  with  temperature  (T).     The  primitive  for  mercury  is 

F  =  .00018179277+175xl0-9772+35116xl0-i07T3. 

Ex.  37.  Interpret  (120)  in  which  e=  instantaneous  counter 
E.M.F.,  L  the  coefficient  of  self-induction,  and  i=  instantaneous 
current;. 

(120)  e~lf 

Ex.  38.  Interpret  (121)  in  which  i  =instantanoeus  current 
flowing  into  a  condenser,  C=  capacity  of  condenser,  e=  instanta- 
neous E.M.F. 

de 


(121) 

^CJt' 

Ex.  39.    Interpret  (122). 

(122) 

1     dt' 

Ex.  4C.  Substitute  (122)  in  (121)  and  give  the  authority  for 
writing  (123). 

(123)  q=Ce. 

Ex.  41.  Interpret  (124)  in  which  $  is  the  magnetic  flux,  E  is 
the  induced  E.M.F.  in  volts,  and  Z  the  number  of  turns  of  wire. 
What  is  the  meaning  of  the  minus  sign? 

Ex.  42.  Construct  a  circle  about  a  center  0  and  draw  its 
radius  OA.    Make  the  radius  /  of  the  circle  about  1.5  ins.  and 


478  PKACTICAL  MATHEMATICS 

incline  it  at  angle  w£  to  the  horizontal.     Draw  a  short  tangent  AD 

(.5  in.)  to  the  circle  from  A,  to  represent  the  linear  velocity  of  A. 

Through  A  draw  a  vertical  line  CB,  intersecting  the  horizontal 

diameter  at  B.    Draw  a  horizontal  line  through  D  intersecting 

BC  at  C.    The  instantaneous  value  of  the  current  is  represented 

di 
by  AB=i,  and  its  simultaneous  rate  of  change  by  CA  =  — .    The 

at 

maximum  value  of  the  current  =0A  =1.  DA  =7w"=2x/7  =  linear 
velocity  of  the  point  A.  The  rate  of  change  of  current  is  the  pro- 
jection of  DA  on  a  vertical  diameter.  Angles  AOB  and  DAC  are 
equal  by  construction.  Prove  (125)  and  (126)  by  applying  the 
definitions  of  trigonometric  functions: 

(125)  i»J  sin  «*. 

di 

(126)  —  =  w7  cos  (ot. 
kit 

Ex.  43.  The  hysteresis  loop,  shown  in  Fig.  190,  may  be  used 
to  determine  the  work  done  in  magnetizing  a  bar  of  iron.  The 
working  formula  is  expressed  in  (127)  in  which  W  is  the  number 
of  ergs  of  work  done  in  one  cycle  for  a  given  mass  of  iron  whose 
volume  is  V.  The  iron  is  subjected  to  a  magnetizing  force  H  and 
the  corresponding  number  B  of  lines  per  square  centimeter  is  noted. 
These  values  of  H  and  B  are  plotted,  giving  the  loop  in  Fig.  190. 
At  the  beginning  of  the  magnetization  the  flux  is  zero,  which  is 
shown  by  the  extra  curve  starting  at  the  origin. 


V   fB\ 


(127)  W=~         HdB. 

^JBt 

The  magnetic  flux  is  reversed  in  direction  and  its  positive  and 
negative  limiting  values  are  designated  by  Bi  and  B2  respectively. 
The  integral  is  identified  with  the  areal  formula  (71),  in  which  y 
is  replaced  by  H  and  dx  is  replaced  by  dB.  Therefore  the  integral 
factor  in  (127)  represents  the  area  measured  between  the  hysteresis 
loop  and  the  Y  axis,  i.e.,  the  summation  of  elementary  strips 
represented  by  HdB.  Since  the  loop  is  bounded  by  two  curves, 
then  the  area  of  the  loop  is  obtained  by  subtracting  the  respective 
areas  under  the  two  curves.  The  same  principle  may  be  applied 
to  obtain  the  difference  between  the  respective  areas  when  they 
are  measured  from  the  X  axis.  Integrate  graphically  the  areas 
under  the  two  bounding  curves  of  the  right  half  of  the  hysteresis 
loop.    A  horizontal  base  fine  passing  through  the  lower  Y  inter- 


RATES,  DERIVATIVES,  AND  INTEGRALS         479 

cept  of  the  loop  serves  as  a  convenient  reference  line  from  which 
to  measure  the  areas.  Trace  the  hysteresis  loop  on  squared 
paper  and  check  the  area  by  counting  squares. 

Ex.  44.  Construct  a  sine  curve  and  label  it  A.  Construct  a 
sine  curve  B  which  leads  A  by  90°  and  whose  amplitude  equals 
x  times  the  amplitude  of  A.    Construct  a  sine  curve  D  lagging 

90°  behind  A  and  whose  amplitude  is  —  times  the  amplitude  of  A. 


Magnetizing  Force 
Fig.  190. — A  Hysteresis  Loop. 

Show  that  equations  (128),  (129),  and  (130)  correspond  to  B,  A, 
and  D  respectively. 

di 
(128)  —  =  x/cosx£. 


dt 


(129) 
(130) 


i=I  sinxZ 


dq 
di' 


COS  xt. 


480  PEACTICAL  MATHEMATICS 

Construct  S  in  phase  with  B  but  whose  amplitude  is  L  times 
that  of  B.  Construct  T  the  negative  of  S.  Show  that  Eqs.  (131) 
and  (132)  correspond  to  S  and  T  respectively. 

(131)  .«-*£. 

(132)  *'-.-'*  a- 

Construct  U  in  phase  with  A  but  whose  amplitude  is  R  times 
that  of  A.    Show  that  (133)  corresponds  to  U. 

(133)  e=Ri. 

Construct  V  in  phase  with  D  but  whose  amplitude  is  —  times 

that    of  D.    Construct  X  the  negative  of  P.     Show  that  (134) 
and  (135)  correspond  to  V  and  X  respectively. 

(134)  «,-* 

(135)  e«'=-<7- 
nzn  •    dq    ndec 

(136)  l=itrcTt' 

Show  that  (136)  may  be  obtained  from  (134)  and  (129)  and 
show  that  (136)  corresponds  to  A. 

Ex.  45.  Construct  the  resultant  of  U  and  S  and  write  the 
corresponding  equation.  Label  the  curve  G  and  let  its  instantaneous 
value  be  represented  by  ex. 

Ex.  46.  Construct  the  resultant  of  U  and  V.  Write  the  corre- 
sponding equation.  Label  the  curve  /  and  let  its  instantaneous 
value  be  represented  by  e2. 

Ex.  47.  Construct  the  resultant  of  U,  S,  and  V.  Write  the 
corresponding  equation.  Label  the  curve  K  and  let  its  instanta- 
neous value  be  represented  by  e3. 

Ex.  48.  Construct  the  graph  for  decaying  currents  from  (137) 
and  show  that  its  derivative  is  expressed  by  139.  Construct  the 
derivative  and  verify  its  ordinates  by  substitution  in  (139). 


RATES,  DERIVATIVES,  AND  INTEGRALS        481 


(137) 


(139) 


t-Jt 


Rt 

L 


(138)      loget=loge/ 


m 

L' 


m 

L' 


Data. 
R=SQ,    L  =  .04  henry. 

7=36.7  amps. 


80 


:>-. 


&  20 


15 


10 


1=36.7 

A.     R  = 

Jfi      L= 

.04  h. 

" — i 

0  12345678 

t 

Time  in  Hundredths  of  Sees. 

Fig.  191. — Decaying  Current  Curve. 

Ex.  49.  Construct  the  graph  for  growing  currents  from  (140) 
and  show  that  its  derivative  is  expressed  by  (141).  Construct  the 
derivative  and  verify  its  ordinates  by  substitution  in  (141).  Show 
that  the  ordinates  of  (140)  may  be  obtained  by  subtracting  the 
ordinates  of  (137)  from  I.    Use  the  data  given  in  Ex.  48. 


(140) 
(141) 


-,(.-.-*). 


di    Ri 
dt=T' 


482 


PEACTICAL  MATHEMATICS 


Ex.  50.  The  growth  and  decay  of  current  in  an  inductive 
circuit  is  expressed  in  (142)  in  which  e  and  i  are  instantaneous 
values  of  E.M.F.  and  current  respectively,  R  is  the  resistance, 
and  L  the  inductance  of  the  circuit.  The  impressed  E.M.F.  e 
is  used  for  two  purposes.      State  which  term  represents  the  part 


30 


25 


15 


10 


\ 

E=110 

V.      R  = 

3fi      L= 

.04  h. 

12  3  4  5  6? 

t 

Time  in  Hundredths  of  Sees. 

Fig.  192. — Growing  Current  Curve. 


of  the  E.M.F.,  corresponding  to  the  resistance  drop  and  which  term 
represents  the  part  of  the  E.M.F.  which  causes  the  current  to 
increase.  Let  e  become  zero,  then  show  that  the  self-induced 
E.M.F.  is  expressed  in  (141) 


(142) 


e=iR+L%. 
at 


CHAPTER  XXIV 
EXPERIMENTAL  CURVES 

1.  In  the  study  of  the  operation  of  electrical  machinery 
and  devices  it  is  customary  to  determine  their  character- 
istics by  observing  their  performance  during  a  series  of 
tests.  Simultaneous  measurements  are  made  of  speed, 
voltage,  armature  and  field  current,  torque,  mechanical 
output,  and  a  series  of  characteristic  curves  are  plotted 
from  the  data.  Owing  to  the  inaccuracy  of  observation, 
errors  of  instruments,  and  fluctuation  of  power,  the  plotted 
data  may  show  slight  deviations.  A  smooth  curve  is  drawn 
through  the  average  of  the  points,  thus  minimizing  the  inac- 
curacies. 

2.  The  Effect  of  the  Variation  of  Load  and  Terminal 
Voltage  on  the  Operation  of  a  Shunt  Motor.  In  Fig.  193 
curve  I  shows  the  relation  between  the  total  amperes  and 
speed  of  a  shunt  motor  which  is  operated  with  a  constant 
impressed  voltage  but  with  variable  load.  Curve  I  shows 
that  the  speed  of  a  shunt  motor  decreases  with  an  increase 
of  load. 

In  Fig.  193  Curve  II  shows  the  relation  between  the 
total  amperes  and  counter  E.M.F.  The  counter  E.M.F. 
is  the  difference  between  the  impressed  voltage  and  the  RI 
drop  in  the  armature.  Curve  II  should  have  about  the 
same  shape  as  I,  since  the  counter  E.M.F.  decreases  with 
decrease  in  speed  and  increase  of  RI  in  the  armature. 

In  Fig.  194  Curve  I  shows  the  relation  between  terminal 
volts  and  speed,  curve  II  shows  the  relation  between  ter- 

483 


484 


PRACTICAL  MATHEMATICS 


minal  volts  and  field  current,  and  curve  III  show  the  relation 
between  terminal  volts  and  current  in  the  armature. 

In  Fig.  (194)  Curve  I  shows  that  the  speed  decreases 
with  decrease  in  terminal  voltage.  Curve  II  shows  that  the 
field    current   varies   directly    with   the   terminal   voltage. 

2000  1(50 


1750 


1500 


1250 


P5 
fllOOO 


750 


500 


250 


150 
140 
130 
120 
110 

100  2 

o 

> 

s 


90 


70  » 

a 

60  | 
50 


Westinghouse  Motor,  10H.P. 

Relation  between  Total  Amperes  and  Speed 
I         "  "  >•  k         >i    Counter  EMF. 


0        5       10      15      20253035      40     455055       6065 
Total  Current  in  Amperes 

Fig.  193. — The  Effect  of    a  Varying    Load    on   the  Operation  of    a 
Shunt  Motor. 

Curve  III  shows  that  the  armature  current  decreases  with 
the  lowering  of  the  terminal  voltage  until  the  knee  of  the 
curve  is  reached  beyond  which  the  current  increases  with  the 
terminal  voltage,  owing  to  the  more  rapid  decrease  in 
counter  E.M.F. 


EXPERIMENTAL  CURVES 


485 


3.  Characteristic  Curves  of  a  Compound  Generator. 
Curves  I,  II,  III,  in  Fig.  195,  show  the  results  of  a  pre- 
liminary test  on  a  compound  generator.  I  was  obtained 
when  the  machine  was  operated  with  the  series  field  cut 

Field  Current  in  Amps. 
*  Armature  Current  in  Amperes 

0123456789       10     11 


150 

140 

ISO 

U0 

110 

kk 
i 

|   00 

> 

\  80 


Westinghouse  Motor  *4325, 10  H.P. 

I       Relation  between  Terminal  Volts  and  Speed 
II  «•  ♦«  "  ♦«       "   Field  Current 

III.  •«  •«  ««  ««       <«  Amperes  Armature 


a  to 

E 

H  60 

50 
40 
30 
20 
10 

°   100  200  300  400  500  600  700  800  900  1000  1100 
Speed  in  R.P.M. 

Fig.  194. — The  Effect  of  Variable  Terminal  Voltage  on  the  Operation 
of  a  Shunt  Motor. 


out  and  therefore  I  shows  the  usual  characteristics  of  a 
shunt  generator.  II  was  obtained  when  the  series  coil  was 
connected  in  cumulatively,  i.e.,  so  that  the  magnetic  field 
of  the  series  coil  was  added  to  the  magnetic    field    of  the 


486 


PRACTICAL  MATHEMATICS 


shunt  coil.  Ill  was  obtained  when  the  series  coil  was 
connected  differentially,  i.e.,  so  that  the  magnetic  field  of 
the  series  coil  was  subtracted  from  the  magnetic  field  of 
the  shunt  coil.  Explain  the  causes  for  the  changing 
voltages  under  I,  II,  and  III  in  Fig.  195. 


140 
130 

_  — 

120 

110  < 

100 

\ 

90 

«80 

\ 

\ 

o 

\ 

OS 

\ 

8 

\ 

I  60 

\ 

B 

3 

H50 

\ 

\ 
\ 

40 

\ 
\ 

i 

Preliminary 

I    Plain  Shunt  Connections 

30 

\ 

II  Cumulative  Compound' « 

m\ 

III  Differential        •«          « 

20 

i 
i 

(Series  field  reversed) 

10 

k 

0      10       20      30      40      59      60      70     80 
Amperes  Line 

Fig.  195. — The  Preliminary  Test  of  a  Compound  Generator. 


In  Fig.  196  a,  b,  c  are  characteristic  curves  of  the  com- 
pound generator,  corresponding  to  II  in  Fig.  195.  They 
show  the  relation  between  terminal  voltage  and  line  current 
when  starting  with  normal  voltage,  90  per  cent  of  normal 
voltage  and  10  per  cent  above  normal  voltage  respectively. 
Explain  by  means  of  the  different  flux  densities  why  the 
voltage  is  nearly  equal  at  full  load  in  the  three  cases. 


EXPERIMENTAL  CURVES 


487 


t  4.  Characteristic  Curves  of  a  Series  Generator.  The 
external  characteristic  of  a  generator  is  the  curve  which 
expresses  the  relation  of  the  terminal  voltage  to  the  line 
current  and  is  represented  by  I  in  Fig.  197.  The  drop 
in  the  armature  plus  the  drop  in  the  field  is  represented 


140 

130 

120* 

{ClU 

fpz*—^ 

110< 

^y 

100, 

yifiy 

I90 

•+j 

■3  80 

> 

I70 

§60 

H 

50 

40 

External  Characteristic 

30 

Curves 

of  General  Electric  Generator 

20 

(a)  Starting  witknormal  voltage 

(b)       •           «     90*    • 

(C)       a           u    10^above«  « 

10 

0       10     20      30      40      50 
Amperes  Line 


70     80 


Fig.  196. — The  Effect  of  Normal,  Subnormal,  and  Abnormal  Initial 
Terminal  Voltage  of  a  Compound  Generator. 

by  the  line  0b  whose  slope  tan  cj)  equals  the  sum  of  the 
armature  and  field  resistances.  Refer  to  paragraph  4 
on  page  305  and  Fig.  115.  The  total  characteristic  curve 
which  expresses  the  relation  between  the  total  generated 
voltage  and  total  current  is  shown  by  curve  II,  Fig.l97,whose 
ordinates  represent  the  sum  of  the  simultaneous  ordinates 
of  I  and  0b.    The  line  Oa  has   a   slope  tan  0  which   equals 


488 


PRACTICAL  MATHEMATICS 


the  sum  of  the  armature  field  and  external  resistances. 
When  the  external  resistance  reaches  a  critical  value  Oa 
will  intersect  I  in  the  neighborhood  marked  m.  Under 
this  condition  the  machine  is  unstable  and  if  the  external 
resistance  exceeds  the  critical  value,  the  machine  cannot 


0       10      20      30    40       50     00       70      80     90       100 
Amperes 

Fig.  197. — The  Characteristics  of  a  Series  Generator. 


build  up.    Why  does  I  rise  to  the  knee  and  then  fall  off 
in  value. 

5.  The  Changes  in  the  Current,  Voltage,  and  Density  of 
the  Electrolyte  of  a  Storage  Battery  during  Discharge.  The 
light    line   shows  the  change  in  current,  the  medium  line 


EXPERIMENTAL  CUEVES 


489 


shows  the  change  in  voltage,  and  the  heavy  line  shows  the 
change  in  the  density  of  the  electrolyte  at  successive  intervals 
of  time.  The  rapid  initial  fall  in  potential  is  due  to  the 
IR  drop  of  the  battery,  showing  that  on  open  circuit  a  storage 
battery  will  indicate  a  higher  voltage  than  will  be  indicated 
immediately  after  closing  the  circuit. 


O Current,  Time 

— — o— —  Density,  Time      Curves 

3 

1.5 

-0 Voltage,  Time 

24 

2.75 
2.5 

1.375 
1.25 

O                      1,             O                       n 

22 

20 

2.25 

1.125 

18 

2 

< 
1 

\ 

16 

1.75 

So 

fl  1.5 
o 
> 
1.25 

.875 
.75 
.625 

9 

ft 

14 1 

12  8, 

s 

10« 

•  1 

.5 

8 

.75 

.375 

6 

.5 

.25 

4 

.25 

.125 

2 

l)        0 

K) 

Time  in  Minutes 

Fig.  198. — The  Discharge  Characteristics  of  a  Storage  Battery. 


6.  The  Efficiency  Test  of  a  Shunt  Motor.  In  Fig.  199  curve 
I  shows  the  relation  between  per  cent  of  no-load  speed  and 
horse-power  output.  The  speed  drops  off  slightly  as  the 
output  of  the  motor  increases.  Curve  II  shows  the  rela- 
tion between  per  cent  efficiency  and  horse-power  output. 
The  efficiency  is  a  maximum  at  full  load. 


490 


PRACTICAL  MATHEMATICS 


In  Fig.  200  curve  III  shows  the  relation  between  torque 
and  horse-power  output.  Curve  IV  shows  the  relation 
between  current  input  and  horse-power  output. 

7.  Comparison  of  the  Starting  Torque  of  Shunt  and 
Series  Motors.  The  light-lined  curve,  Fig.  201,  shows  the 
torque  of  a  series  motor  which  at  starting  is  greater  than 
the  torque  of  shunt  motors  whose  curves  are  represented 


H.P.  Output 
Fig.  199. — The  Efficiency  Curve  of  a  Shunt  Motor. 


by  the  medium  and  heavy  lines.     How  is  the  difference 
explained? 

8.  Fig.  202  shows  the  performance  of  a  rotary  converter 
when  generating  from  D.C.  to  A.C.  The  light  line  shows 
the  relation  between  D.C.  voltage  and  output.  The  heavy 
line  represents  the  relation  between  A.C.  voltage  and  out- 
put. The  dotted  line  represents  the  relation  between 
efficiency  and  the  output. 


EXPERIMENTAL  CURVES 


491 


Ex.  1.  A  shunt  motor  at  rest  shows  a  field  current  of  4.1  amps, 
when  112  volts  are  applied  to  its  terminals  and  the  armature 
shows  a  drop  of  .15  volt  under  the  +  and  —  brushes  when  a 
current  of  5  amps,  flows  through  it.  Determine  the  field  resist- 
ance and  the  armature  resistance  of  the  motor. 

Ex.  2.  Determine  the  power  curve  for  the  discharge  of  the 
storage  batteries  whose  performances  are  illustrated  in  Fig.  198. 
Integrate  the  power  curve  and  obtain  the  watt  hour  curve. 


4        5 
H.P.  Output 

Fig.  200.— Efficiency  Test  of  a  Shunt  Motor. 

Ex.  3  Determine  the  electrical  efficiency  curve  of  the  shunt 
motor  whose  performance  is  illustrated  in  Fig.  194. 

Ex.  4.  Fig.  203  represents  a  transverse  section  through  the  field 
frames  of  a  multipolar  shunt  generator.  Fig.  204  represents  a 
longitudinal  section  through  the  shaft  and  a  pair  of  opposite  poles 
of  the  same  machine.  Study  a  similar  generator  enumerating  the 
following  numbered  parts  and  add  their  respective  dimensions 
together  with  a  description  of  the  kind  of  materials  used  in  their 


492 


PRACTICAL  MATHEMATICS 


construction.  Determine  the  weight  of  each  detachable  part 
wherever  possible  and  estimate  the  parts  which  cannot  be  dissem- 
bled. 

Weigh  the  entire  machine.     The  following  list  identifies  the 
numbered  parts:    (1)  base;    (2)    field  frame  or  yoke;    (3)  field 

G.E.  Type  CE-Form  A  Motors 
1050  R.P.M.  115  Volts-156  Amps-2  H.P. 


0      2.5 


7.5      1U     12.5      15     17.5     20     22.5    25 
Amperes  Armature 


Fig.  201. — Starting  Torques  of  Series  and  Shunt  Motors. 


cores;  (4)  series  field  coils;  (5)  shunt  field  coils;  (6)  pole  shoes; 
(7)  bearing  pedestals;  (8)  armature;  (9)  armature  coils;  (10) 
armature  core;  (11)  armature  spider;  (12)  air-gap;  (13)  air  ducts; 
(14)  shaft;  (15)  commutator;  (16)  insulation;  (17)  brushes; 
(18)    holders;     (19)    rocker  arm;.  (20)    pulley.     Supplement   the 


EXPERIMENTAL  CURVES 


493 


above  data  with  complete  information  regarding  the  manufacturer 
and  the  type  of  the  machine;  its  rated  capacity,  volts,  amperes, 
number  of  poles,  speed,  per  cent  of  polar  span,  both  maximum 
and  minimum,  size  of  wire,  weight  per  spool,  and  number  of  turns 
of  both  shunt  and  field  coils,  length  and  diameter,  and  volume  of 
armature  core,  thickness  of  laminations,  number  of  ventilating 

140 


ISO 

120 

i 

110 


70 


i;o 


50 


40 


20 


10 


X 


/ 


.j**' 


Westinghouse 
Generator 

Output  in  Kw. 


90 


70  o 

a 

t>603 


50 


40 


30 


20 


10 


a 


0  .25  .50  .75  1  1.25  1.50 

O D.C.  Voltage  and  Output 

O  A.C.  Voltage  and  Output 

— — O ejs,  Efficiency  and  Output 

Fig.  202. — The  Performance  of  a  Rotary  Converter. 

ducts,  number  and  width  of  slots,  number,  width,  depth,  length, 
and  active  length  of  commutator  bars,  thickness  of  mica  insulation, 
volts  per  bar,  style  of  armature  winding,  number  of  coils,  coils 
per  slot,  turns  per  coil,  length  of  one  turn,  weight  of  wire  on 
armature,  number  of  brush-holder  studs,  number  of  brushes  per 
stud,  amperes  per  square  inch  of  contact  surfaces,  brush  pressure. 
With  the  above  data  as  a  suggestive  guide  make  a  complete 
design  of  the  above  machine,  showing  in  detail  all  calculations. 


494 


PKACTICAL  MATHEMATICS 


Fig.  203. — Transverse  Section  of  a  Multipolar  Generator. 
2 


Fig.  204. — Longitudinal  Section  of  a  Multipolar  Generator. 


EXPERIMENTAL  CURVES  495 

Give  an  approximate  estimate  of  its  operation  under  various  con- 
ditions of  load.     Make  a  sketch  of  the  machine  in  cross-section. 

9.  Electrical  Measuring  Instruments.  Station  and 
laboratory  measuring  instruments  may  be  classified  under 
one  of  the  following  heads  according  to  the  principle  of 
their  construction:  (a)  Electrodynamic;  (6)  electromag- 
netic; (c)  magnetodynamic;  (d)  electrothermic;  (e)  electro- 
static; (/)  resonating.  These  instruments  are  used  for 
measuring  current,  voltage,  resistance,  conductivity,  power, 
frequency,  power  factor,  and  also  the  wattless  component 
of  the  current.  The  magnetodynamic  type  of  instrument 
can  be  used  for  direct  current  measurements  only,  whereas 
all  other  types  may  be  used  both  for  direct  and  alternating 
current  measurements.  Instruments  are  further  classified 
as  indicating,  recording,  or  integrating  types.  Indicating 
instruments  possess  an  index  or  movable  pointer  whose 
position  on  a  graduated  scale  indicates  the  magnitude  of 
of  the  quantity  which  is  measured.  A  recording  instru- 
ment has  its  index  or  movable  pointer  adjusted  with  a 
stylus  or  inking  point  under  which  a  strip  or  disc  of  paper 
is  moved  uniformly  so  as  to  produce  a  continuous  record. 
Integrating  instruments  possess  a  clock  mechanism  which 
propels  a  series  of  pointers  or  indexes  over  graduated  dials. 
The  dials  are  similar  to  the  dials  of  gas  meters  and  water 
meters. 

10.  Siemens  Dynamometer.  The  construction  of  a 
Siemens  .  dynamometer  is  shown  in  Fig.  205.  It  is  an 
electrodynamic  type  and  measures  the  current  strength 
by  the  reaction  between  the  current  flowing  through  the 
fixed  coils  C\  and  C2  and  the  movable  coil  MC.  The  movable 
coil  MC  contains  a  few  turns  of  wire  and  is  suspended  by 
the  thread  F.  Its  terminals  dip  into  the  mercury  cups  X 
and  Y.  A  helical  spring  S  is  attached  to  the  upper  end  of 
MC  and  also  to  the  movable  knob  K.  A  pointer  Pi  is 
attached  to  the  movable  coil  MC  and  its  motion  is  limited 
by  two  stops  placed  on  the  graduated  dial  D.    A  second 


496 


PEACTICAL  MATHEMATICS 


pointer  P  is  attached  to  the  knob  $,  so  that  when  S  is 
turned  P  indicates  a  reading  on  the  circular  scale  of  D. 


c2         Ci 
Electrical  Connections 


Fig.  205. — A  Siemens  Dynamometer. 

The  fixed  coil  is  rigidly  supported  by  a  wooden  standard, 
so  that  its  plane  is  perpendicular  to  that  of  the  movable 


EXPERIMENTAL  CURVES  497 

coil.  The  fixed  coil  has  two  windings  designated  by  C\ 
and  C2  which  are  used  for  strong  and  weak  currents  respect- 
ively. The  dial  is  graduated  into  400  divisions.  The 
electrical  connections  show  that  in  operation  the  movable 
and  fixed  coils  are  connected  in  series  so  that  when 
current  flows  through  them  the  movable  coil  is  deflected. 
The  knob  is  turned  to  bring  the  movable  coil  back  to  its 
normal  position  at  right  angles  to  the  fixed  coil.  The 
force  F  exerted  by  the  spring  is  proportional  to  the  angle  0 
of  torsion,  which  is  read  from  the  position  of  P.  The  force 
exerted  on  the  movable  coil  is  proportional  to  the  square 
of  the  current  I.     Therefore 


Foc6. 

(1) 

F  =  Zi0. 

FccP. 

(2) 

F  =  K2P. 

60c/2. 

(3)     • 

..        6=^/2  =  ^/2, 
Ai 

0i    /12 

02       h2' 

(5) 

7i     Ve7 

(4) 


The  interpretation  of  (5)  states  that  the  current  is 
proportional  to  the  square  root  of  the  angle  through  which 
the  pointer  P  has  been  turned. 

Ex.  5.  What  are  the  constants  of  two  Siemens  dynamometers, 
one  of  which  indicates  220°  of  angular  twist  when  a  current  of  20 
amperes  flows  through  it  and  the  other  indicates  175°  of  angular 
twist  when  17  amperes  flows  through  it. 

11.  The  Thomson  ammeter,  shown  in  Fig.  206,  and  the 
Queen  ammeter,  shown  in  Fig.  205,  are  instruments  of  the 
electromagnetic  type.  In  such  instruments  the  forces 
act  between  the  parts  in  such  a  way  as  to  reduce  the 
reluctance  of  the  system  to  a  minimum.  In  the  Thomson 
instrument  the  current  which  is  to  be  measured  is  passed 
through  the  inclined  cylindric  coil  EE.  The  movable  part 
consists  of  a  vertical  shaft  mounted  between  jewel  bearings. 


498 


PEACTICAL  MATHEMATICS 


A  pointer  B,  a  soft  iron  vane  V,  a  damper  K,  and  one  end 
of  the  spiral  spring  are  attached  to  the  shaft.  The  other 
end  of  the  spring  is  fixed  to  the  support  F  and  in  this  manner 
serves  as  a  controlling  force  to  oppose  the  electromagnetic 


Fig.  206. — Thomson  Ammeter. 


force  between  the  coil  EE  and  the  vane  V,  and  also  serves 
to  restore  the  pointer  to  its  zero  point  upon  the  cessation 
of  current.  In  the  Queen  instrument  the  current  which 
is  to  be  measured  is  passed  through  the  link-shaped  coil  CC. 


EXPERIMENTAL  CURVES 


499 


The  movable  part  consists  of  a  vertical  shaft  mounted 
between  jewel  bearings.  A  V-shaped  piece  of  soft  iron  V, 
which  has  been  bent  into  a  semicylindric  shape,  is  attached 
to  the  shaft  together  with  a  pointer  P,  a  damper  F,  and  a 
spiral  spring  S.  The  other  end  of  the  spring  is  attached  to 
the  support  /  and  serves  as  a  controlling  force  to  oppose  the 
electromagnetic  force  between  the  coil  CC  and  the  vane  V. 
The  field  in  the  ends  of  the  coil  is  a  maximum.  The 
iron  vane  V  tends  to  turn  in  such  a  position  that  more 
lines  of  force  may  pass  through  it. 


Fig.  207. — Queen  Ammeter. 

In  the  Thomson  and  Queen  instruments  the  force  is 
proportional  to  the  square  of  the  flux  density,  and  therefore 
if  the  permeability  of  the  iron  is  constant  the  force  is  pro- 
portional to  the  square  of  the  current.  An  instrument 
which  measures  the  effective  value  of  an  alternating  current, 
i.e.,  the  square  root  of  the  average  square  of  the  instanta- 
neous currents  must  exert  a  torque  which  is  proportional 
to  the  square  of  the  current.  ' 

Factf.  (6)  F  =  Kk5>2. 

cj>oc/.  (7)  $  =  K2I. 

Fez  P.  (8)     /.    F  =  KiK22P  =  KI2. 


500 


PEACTICAL  MATHEMATICS 


The  modification  of  the  electromagnetic  type  is  repre- 
sented by  the  Whitney  D.C.  voltmeter,  shown  in  Fig.  208. 
The  current  which  is  to  be  measured  passes  through  the 
solenoid  CC  which  produces  a  magnetic  field  at  right  angles 
to  the  field  of  the  permanent  magnet  MM.  The  moving 
part  is  a  jewel-pivoted  shaft  to  which  is  attached  the  soft 


Fig.  208.— Whitney  D.C.  Voltmeter. 


iron  vane  V.  The  permanent  magnet  serves  as  the  con- 
trolling force  to  bring  the  pointer  to  zero  after  the  cessation 
of  current.  The  position  of  the  vane  when  the  instrument 
is  operative  depends  upon  the  resultant  of  the  magnetic  field 
due  to  both  MM  and  CC. 

The  Weston  D.C.  ammeter,  shown  in  detail  in  Fig.  209, 


EXPERIMENTAL  CURVES 


501 


represents  the  magnetoelectro  type  and  is  better  known 
as  a  permanent  magnet  instrument.  The  magnetic  field  is 
provided  by  the  permanent  magnet  to  which  a  pair  of  soft 
iron  pole  pieces  BB  are  attached.  A  soft  iron  cylindric  core 
is  centrally  placed  between  the  shaped  pole  pieces  so  as  to 
insure  a  very  strong  and  uniform  field.  The  movable  part 
consists  of  a  jewel-pivoted  aluminum  frame  wound  with  a 
coil  of  fine  wire.     One  spiral  control  spring  is  mounted  above 


Coil  (top) 


U7 

Coil  (side) 


Fig.  209. — Detail  of  Weston  Ammeter. 


the  coil  and  another  spiral  control  spring  is  mounted  below 
the  coil  and  in  addition  these  springs  serve  to  convey  the 
current  from  the  moving  coil  of  the  instrument.  A  pointer 
is  attached  to  the  movement.  This  instrument  has  a 
uniform  scale.  Owing  to  its  simplicity,  its  superior  mechan- 
ical construction,  its  freedom  from  errors  and  its  electrical 
efficiency,  the  Weston  instrument  is  universally  recognized 
as  a  precision  standard  for  the  calibration  of  all  types  of 
instruments.     The  Weston  instrument  is  a  millivoltmeter 


502  PRACTICAL  MATHEMATICS 

and  when  it  is  provided  with  adjustable  internal  resistances 
which  are  placed  in  series  with  its  movable  coil  it  becomes 
a  voltmeter.  When  the  moving  coil  is  connected  in  parallel 
with  a  low-resistance  conductor  called  a  shunt,  the  instru- 
ment becomes  an  ammeter.  Shunts  are  adjusted  or  tapped, 
so  that  they  give  a  drop  of  50  millivolts  when  the  maximum 
current  flows  through  them.  Since  all  ammeters  are 
adjusted  so  as  to  be  operative  over  a  full-scale  deflection 
with  50  millivolts  any  ammeter  may  be  used  interchange- 
ably on  any  shunt.  Extra  external  resistances  called  multi- 
pliers are  used  in  series  with  a  voltmeter  for  the  purpose  of 
increasing  the  range  of  its  measurement.  The  force  at 
each  position  of  the  coil  is  proportional  to  the  current 
flowing  through  the  coil,  since  the  flux  and  number  of  turns 
remain  constant. 

Ex.  6.  Determine  the  resistance  of  a  1000-ampere  shunt  to 
give  a  drop  of  50  millivolts. 

Ex.  7.  The  resistance  of  a  150-volt  Weston  voltmeter  is  14,000 
ohms.  Determine  the  resistance  of  a  suitable  multiplier  which 
when  used  in  series  with  the  voltmeter  will  make  it  possible  to 
measure  a  maximum  voltage  of  600. 

Ex.  8.  A  direct  reading  voltmeter  has  150  uniform  scale  divi- 
sions, and  is  calibrated  for  150  volts.  The  observer  can  approxi- 
mate to  .1  of  a  volt  throughout  the  entire  range.  What  is  the 
percentage  error  due  to  the  error  of  reading,  (a)  when  the  instru- 
ment indicates  100  volts;  (6)  at  150  volts;  (c)  at  10  volts;  at 
1  volt.  Draw  a  curve  showing  the  per  cent  of  error  plotted  verti- 
cally and  the  voltage  horizontally. 

Ex.  9.  A  direct  reading  ammeter  calibrated  with  100  divisions 
reads  100  amperes.  Allow  that  the  observer  can  approximate  to 
.1  of  a  scale  division.  Plot  the  curve  between  per  cent  error  and 
current. 

Ex.  10.  A  portable  voltmeter  is  checked  with  a  standard  instru- 
ment. The  former  reads  respectively:  59.8,  69.7,  79.8,  89.8, 
99.6,  when  the  latter  reads  correspondingly  60,  70,  80,  90,  100 
volts.  What  is  the  per  cent  error  in  the  portable  instrument  for 
the  corresponding  standard  readings  ? 

Ex.  11.  A  Siemens  dynamometer  has  a  uniform  scale  and  it 
is  assumed  that  the  observer  can  approximate  to  \  scale  division. 
Plot  the  curve  for  per  cent  error,  using  the  data  in  Ex.  5. 


EXPERIMENTAL  CURVES  503 

Ex.  12.  In  a  direct  reading  A.C.  instrument  (voltmeter  or 
ammeter)  the  value  of  each  scale  division  is  the  same  throughout 
the  scale,  although  the  scale  is  not  uniformly  spaced.  Suppose 
each  scale  division  represents  one  ampere,  then  in  approximating 
to  tenths  of  a  division  the  error  is  inversely  proportional  to  the 
width  of  the  division.  What  is  the  relative  error  in  reading  100 
amperes  compared  to  10  amperes? 

Ex.  13.  D.C.  instruments  give  deflections  which  are  propor- 
tional to  the  current,  whereas  A.C.  instruments  intended  for  both 
D.C.  and  A.C.  measurements,  give  deflections  which  are  propor- 
tional to  the  square  of  the  current.  Considering  a  150-amp. 
instrument  with  scale  divisions  corresponding  to  1  amp.  each 
what  is  the  relative  per  cent  error  for  an  error  made  at  10  amp. 
and  100-amp.  divisions  respectively? 


CHAPTER  XXV 
INDUCTANCE  AND   CAPACITY 

1.  Counter  E.M.Fs.  Whenever  an  E.M.F.,  e,  is  applied 
to  a  circuit  it  has  two  functions  to  perform,  i.e.,  a  part  e\ 
is  used  to  overcome  the  resistance  of  the  circuit  and  a 
second  part  62  is  used  to  overcome  any  counter  E.M.F. 
which  may  be  set  up  in  the  circuit.  In  a  D.C.  circuit 
counter  E.M.Fs.  are  transient  and  are  observed  for  a  brief 
moment  immediately  after  closing  and  opening  a  circuit. 
In  an  A.C.  circuit  counter  E.M.Fs.  are  continuously  present 
and  constitute  an  appreciable  element  in  the  consideration 
of  the  E.M.F.,  which  is  applied  to  or  impressed  upon  a 
circuit.     This  relation  is  expressed  in  (1). 

(1)  e  =  ei+62. 

(la)  Applied    E.M.F.  =  E.M.F.    to    overcome  resistance+ 
E.M.F.  to  overcome  counter  E.M.F. 

Every  electric  circuit  possesses  three  distinct  properties, 
i.e.,  resistance,  inductance  and  capacity,  whose  respective 
magnitudes  may  range  from  negligible  values  to  infinite 
values.  These  three  properties  are  comparable  to  the 
friction,  inertia,  and  elasticity  of  a  mechanical  circuit, 
which  may  be  illustrated  by  a  fluid  in  motion  or  by  any 
moving  mechanism. 

An  electric  conductor  is  surrounded  by  a  magnetic  field  of 
force  and  also  by  an  electrostatic  field  of  force  owing  to  its 
proximity  to  other  conductors  and  substances  in  nature. 
The  passage  of  an  electric  current  in  a  conductor  sets  up  a 
disturbance  in  the  surrounding  fields  of  force  which  requires 

504 


INDUCTANCE  AND  CAPACITY  505 

a  definite  expenditure  of  electric  energy  in  the  circuit. 
Any  change  in  the  current  which  is  supplied  to  a  circuit 
will  cause  a  simultaneous  change  in  the  energy  which  is 
stored  in  the  surrounding  fields.  The  energy  in  the  fields 
surrounding  a  conductor  carrying  a  direct  current  will 
persist  unchanged  as  long  as  the  current  in  the  conductor 
is  maintained  uniform,  i.e.,  at  a  constant  amperage.  A 
change  in  the  intensity  of  the  current  flowing  through  a 
conductor  produces  a  simultaneous  change  in  the  intensity 
of  the  surrounding  fields.  Again,  any  change  in  the  inten- 
sity of  the  surrounding  field  reacts  upon  the  conductor  and 
induces  or  produces  therein  a  counter  E.M.F.  of  induct- 
ance when  it  is  due  to  the  magnetic  field  and  a  counter 
E.M.F.  of  capacity  when  it  is  due  to  the  electrostatic 
field. 

2.  Counter  E.M.F.  of  Self-induction.  The  counter 
E.M.F.  of  self-induction  impedes  the  introduction,  varia- 
tion, and  extinction  of  a  current  passing  through  a  con- 
ductor and  in  a  D.C.  circuit  gives  rise  to  the  phenomena 
of  growing  and  decaying  currents  as  illustrated  in  Figs.  192 
and  191  respectively.  Therefore  a  momentary  but  definite 
time  elapses  in  a  D.C.  circuit  until  an  increasing  or  growing 
current  reaches  its  maximum  value  and  a  momentary  but 
definite  time  elapses  in  a  D.C.  circuit  until  a  decaying  or 
falling  current  vanishes.  These  transient  effects  are  pro- 
portional to  the  counter  E.M.F.,  ex,  of  self-induction  which 
is  in  turn  proportional  both  to  the  rate  of  change  of  the 
magnetic  field  set  up  by  the  conductor,  and  to  the  number 
N  of  turns  of  conductor  linked  with  the  changing  field  as 
expressed  in  (2). 

(2)  *•*$ 

Observation.  A  counter  E.M.F.  of  one  volt  is  induced  in  a 
circuit  when  the  number  of  lines  of  force  passing  through  or 
linked  with  it  changes  at  the  rate  of  108  per  second. 


506  PRACTICAL  MATHEMATICS 

In  a  D.C.  circuit  after  the  transient  effect  has  disappeared 
the  E.M.F.  impressed  upon  the  circuit  overcomes  resistance 
alone  and  therefore  the  resulting  continuous  current  is 
expressed  in  (3),  but  more  precisely  by  (3a) : 

(3)  *=§.  (3a)  7=|. 

If  an  alternating  E.M.F.,  e\  be  impressed  upon  the 
above  circuit  the  resulting  current  cannot  attain  the  same 
maximum  value  I  of  the  direct  circuit,  as  the  delayed 
current  would  tend  to  fall  as  soon  as  the  E.M.F.  is  reversed. 
If  the  frequency  is  increased  the  rate  of  change  of  current 
is  increased  which  means  a  more  rapid  change  of  flux  and 
consequently  a  greater  counter  E.M.F.  and  therefore  a 
reduced  maximum  value  which  can  be  attained  by  the 
alternating  current. 

Observation.  Inductance  is  a  property  of  an  electric 
circuit  which  causes  a  counter  E.M.F.  to  be  set  up  in  it  when- 
ever a  change  of  current  takes  place  in  it. 

3.  Unit  of  Inductance.  A  circuit  or  piece  of  apparatus 
has  one  unit  of  inductance  called  a  henry  (L)  when  a  counter 
E.M.F.  of  one  volt  is  set  up  in  it  by  a  current  which  changes 
at  the  rate  of  one  ampere  per  second: 

dt 

In  a  D.C.  circuit  the  impressed  E.M.F.  is  expressed  in 
(6),  which  results  from  substituting  in  (1)  the  value  of  e\ 
from  (3)  and  the  value  of  62  from  (5).  The  positive  sign 
is  placed  before  eg,  since  it  overcomes  a  counter  E.M.F. 
which  is  represented  by  a  negative  sign,  as  shown  in  (5). 

di 
(6)  e=iR+L 


dt' 


(7)  e~iR=Lft. 


INDUCTANCE  AND 

CAPACITY 

—  -rdt-- 

Li 

di 
e' 
l~R 

-%*• 

'ft 

di 
e 
~R 

RtV 

=  log  (i- 

~R/\o 

*  = 

#- 

507 


(8) 
(9) 

(10) 

(11) 

In  (9)  time  extends  from  zero  to  t  seconds  and  the 
current  ranges  from  0  to  i  amperes.  Substituting  these 
limiting  values  in  (10)  and  writing  the  result  with  exponen- 
tial notation  we  obtain  (11),  which  is  the  equation  of  a 

L 

growing  current.     The  ratio  —  is  called  the  time  constant 

H 

of  the  circuit. 

The  equation  for  a  decaying  current  may  be  obtained 

from  (1)  by  considering  the  condition  of  the  circuit  at  the 

instant  of  removing  the  impressed  E.M.F.,  i.e.,  when  e  =  0, 

as  expressed  in  (12) : 

(12)  e  =  o  =  iR+Ljt. 

(13)  -|d*=* 


L         i 


(14) 
(15) 


-ff*.fft 

LJ0  Je_     I 

R 

T    .    T 


Rt 
L 


R 
Rt 

R        ■ 


(16) 

(16)  is  recognized  as  the  equation  of  a  decaying  current. 


508  PRACTICAL  MATHEMATICS 

Ex.  1.  Determine  the  inductance  of  the  following  mirror 
galvanometers:  (a)  resistance  =  5000  ft  and  time  constant  = 
.0004  sec;  (b)  resistance  =  2700 ft  and  time  constant  =  .001  sec; 
(c)  resistance  =  1,000,000  ft  and  time  constant  =  .0007  sec 

4.  In  an  alternating  current  circuit  the  counter  E.M.F. 
62  may  be  expressed  in  (18)  by  differentiating  (17)  and 
substituting  in  (12) 

(17)  i  =  Isirn)Jt. 

(18)  62  =  —  w/L  cos  0)2. 

62  is  a  maximum  when  o>2  =  0,  as  shown  in  (19),  but 
when  oit  =  0,  then  i  =  0 : 

(19)  62=  —  w/L  (maximum). 

Therefore  the  counter  E.M.F.  62  of  self-induction  lags 
90°  behind  the  current  and  in  order  to  overcome  the 
counter  E.M.F.  62  the  impressed  E.M.F.  e  must  have  a 
component  6r2  equal  and  opposite  to  62  which  leads  the 
current  by  90°: 

(20)  ef2  =  L-r  =  oIL  cos  at. 

(21)  e'2  =  uIL  (maximum). 

(22)  ^  =  o)L  =  Xz, 

The  maximum  value  of  e'2  is  expressed  in  (21)  which 
occurs  when  o)2  =  0  or  when  i  =  0.  (22)  results  from  dividing 
(21)  by  /.  The  right-hand  member  of  (22)  toL  is  abbre- 
viated XL,  which  is  called  the  inductive  reactance.  From 
(21)  XL  may  be  defined  as  the  factor  which  is  multiplied 
into  the  current  to  give  6r2  the  component  of  the  impressed 
E.M.F.,  6,  which  overcomes  the  counter  E.M.F.,  62,  of  self- 
induction. 

Ex.  2.     Describe  and  illustrate  the  graphs  of  e,  e\,  e2,  e'2,  and  i. 
Ex.  3.     Construct  (11)  and  (16)  on  semilog  paper. 


INDUCTANCE  AND  CAPACITY  509 

5.  The  inductance  of  a  long  solenoid  is  expressed  in 
(23)  in  which  L  =  inductance  in  henrys,  N  =  number  of 
turns,  [l  =  permeability  of  core,  A  =  section  area  of  the  core 
in  square  centimeters,  1  =  length  of  core  in  centimeters, 
r=mean  radius  of  the  coil  in  centimeters: 

L26JVW 
(23)  L-       Wl      . 

Show  that  (23)  reduces  to  (24)  for  a  circular  section: 
4x2iWu. 


(24)  L  = 


109Z 


Ex.  4.  Determine  the  inductance  of  the  primary  coil  of  a 
transformer  having  500  turns.  The  iron  core  is  70  cm.  in  length 
and  has  a  cross-section  area  of  325  sq.cm.  Assume  u.  constantly 
equal  to  1400.  Draw  the  curves  of  induced  voltage  for  both  25 
and  60  cycles  when  the  effective  primary  current  is  10  amps. 

6.  The  inductance  in  henrys  of  a  length  I  of  straight 
cylindric  wire  of  radius  r  and  permeability  [l  surrounded 
by  an  outside  medium  of  permeability  one  is  expressed  by 
the  approximate  formula  (25) : 

(25)  L  =  J9[logf2j-l+!]. 

The  inductance  in  henrys  of  a  return  circuit  of  two 
parallel  wires  each  of  length  I  is  expressed  in  the  approx- 
imate formula  (26),  wherein  d  is  the  distance  between  wires: 


(26) 


=  1041Og*r+4-T} 


7.  Mutual  Inductance.  A  current  flowing  in  one  con- 
ductor will  induce  an  E.M.F.  and  a  corresponding  current 
in  a  neighboring  conductor  which  is  not  electrically  connected 
with  it.  The  phenomenon  of  mutual  inductance  is  similar 
to  the  phenomenon  of  self-inductance.  Mutual  inductance 
M  is  measured  in  henrys. 


510  PRACTICAL  MATHEMATICS 

(27)  expresses  mutual  inductance  of  two  coils  or  solenoids 
wound  on  the  same  core  and  of  approximately  equal 
diameters.  JVi  and  N2  are  the  respective  number  of  turns 
on  the  solenoids,  I  the  total  or  combined  length  of  the 
solenoids  in  centimeters,  [x  =  the  permeability  of  the  core, 
and  r  the  radius  of  the  core  in  centimeters. 

Comparing  (27)  with  (24),  we  observe  that  the  only 
difference  is  the  substitution  of  N1N2  for  N2. 

Ex.  5.    From  (24)  and  (27)  show  that  Mm \rLjZ. 

Ex.  6.  Determine  the  mutual  inductance  of  two  coils  of  110 
and  1100  turns  respectively  wound  on  the  core  of  Ex.  2.  In  (27) 
xr2  may  be  replaced  by  the  sectional  area  A  of  the  core. 

The  mutual  inductance  of  two  parallel  wires  of  length  I,  radius  r, 
and  at  a  distance  d  apart  is  expressed  in  the  approximate  for- 
mula (28). 

(28)  *=#««f-W# 

Ex.  7.  Transform  (25)  and  (28)  so  as  to  express  L  and  M 
with  common  logarithms. 


Ex.  8.     Solve  (25)  for 


(29)     ^+1  -7-=log€ Mult.,  sub.  in  (25) 


21  4         °    r 


(30) 


!^-#+I-i] ^"» 


21 

(31)  r  = r-    9  =r Inversion  of  (30) 

logn-2r+1-iJ 

(32)  -•=  e  2l  4  Def .  log  (29) 

r 

(33)  r  =2Ze    L  21  4  J Inversion  of  (32) 


INDUCTANCE  AND  CAPACITY  511 

The  values  of  r  are  expressed  in  (31)  and  (33),  the  former  being 
the  antilogarithmic  form  and  the  latter  the  exponential  form. 
The  shape  of  the  graphs  of  (33)  is  illustrated  in  Figs.  144,  171, 
and  191. 

8.  Practical  Values  of  Inductances.  A  pair  of  line 
wires  strung  on  a  telephone  pole  will  have  two  to  four 
milhenrys  per  mile,  according  to  the  displacement  of  the 
wires.  The  resistance  of  induction  coils  varies  from  6000 
to  80,000  ohms  and  have  a  corresponding  inductance  of 
50  to  2000  henrys. 

Observation.  Inductance  is  a  property  of  an  electric 
circuit  which  manifests  itself  in  a  counter  E.M.F.  whenever  a 
change  is  made  in  the  strength  of  the  current  flowing  through  a 
conductor  or  when  a  change  is  made  in  the  strength  of  the 
magnetic  field  surrounding  a  conductor.  Inductance  may  be 
distributed  along  a  circuit  or  it  may  be  concentrated  in  a  coil 
called  a  reactor. 

Ex.  9.  The  inductance  of  a  field  spool  of  a  generator  is 
expressed  in  (34)  wherein  (J>  is  the  total  flux  from  one  pole,  n  the 
number  of  turns  per  spool,  //  the  field  current  of  the  machine. 

(34)  L-* 


108J, 


What  is  the  inductance  of  a  bi-polar  generator  having  3000 
turns  per  spool  and  field  current  of  2.5  amps,  with  a  flux  of  2  mega- 
maxwells? 

Ex.  10.  What  is  the  inductance  of  an  anchoring  of  cast  steel 
wound  with  200  turns  of  wire  carrying  5  amps.  The  mean  diameter 
is  5  ins.  and  the  radius  of  cross-section  1  in. 

Ex.  11.  Determine  the  self-induction  of  a  solenoid  consisting 
of  ten  layers  of  No.  16  double  cotton  covered  wire  wound  upon 
a  cylindrical  core  1  in.  diameter.  Compute  for  wood  as  well  as 
for  a  soft  iron  core. 

9.  Capacity.  Two  neighboring  conductors  or  a  con- 
ductor and  the  earth  constitute  an  electric  pair  when  they 
are  not  in  electric  connection.  Such  a  pair  is  surrounded 
and  separated  by  an  electrostatic  field  and  any  change  in 


512  PRACTICAL  MATHEMATICS 

the  electric  condition  of  either  member  of  the  pair  pro- 
duces a  corresponding  change  in  the  electrostatic  field. 
The  latter  in  turn  reacts  upon  the  other  member  of  the  pair 
to  change  its  electric  condition.  The  intensity  with  which 
one  member  of  a  pair  influences  the  potential  and  the 
charge  upon  the  other  member  is  measurable  and  this 
property  of  a  circuit  is  called  its  capacity.  The  electro- 
static field  behaves  like  an  elastic  medium  and  therefore 
the  term  tension  is  applied  to  it.  It  may  be  likened  to 
the  pistons  which  separate  the  fluids  in  the  pipe  system, 
shown  in  Fig.  210  A.  It  requires  a  definite  quantity  of 
electricity  to  charge  an  electric  circuit,  i.e.,  to  create  a  poten- 
tial difference  between  an  electric  pair  and  therefore  a  definite 


Fig.  210A. — A  Mechanical  Fig.  210B. — Diagrammatic  Repre- 

Condenser.  sentation  of  a  Condenser. 

expenditure  of  energy  in  order  to  increase  the  tension  in 
the  electrostatic  field  surrounding  a  conductor. 

Capacity  may  be  distributed  throughout  a  circuit  or 
it  may  be  concentrated  or  condensed  in  a  special  device 
called  a  condenser  or  storage  pocket,  shown  diagramatically 
in  Fig.  2105.  A  condenser  is  formed  by  spreading  the 
exposed  surfaces  of  two  insulated  conductors  and  at  the 
same  time  reducing  the  distance  between  them.  Two  long 
and  broad  sheets  of  metal  would  make  a  cumbersome 
device,  but  this  apparent  disadvantage  is  overcome  by 
winding  the  insulated  strips  into  a  bundle  like  a  bolt  of 
cloth.  Another  method  is  to  divide  the  long  strips  into 
leaves  which  are  interlaced  so  that  the  two  sets  of  alternate 
leaves  are  connected  to  the  two  respective  conductors. 
As  a  consequence  of  its  charge  and  voltage  a  condenser 


INDUCTANCE  AND  CAPACITY  513 

manifests  itself  as  a  counter  E.M.F.     Its  ability  to  take  a 
charge  of  electricity  is  called  its  capacity. 

The  capacity  of  a  condenser  is  a  numeric  quantity  and 
equals  the  magnitude  of  the  charge  which  produces  a  unit 
difference  of  potential  between  its  pair  of  elements.  A 
condenser  whose  difference  of  potential  is  raised  one  volt, 
E,  by  a  charge  of  one  coulomb,  Q,  has  one  farad,  F,  i.e.,  one 
unit  capacity.  A  farad  is  10  ~9  times  the  absolute  unit. 
A  subsidiary  unit  which  is  more  convenient  in  practice  is 
one  microfarad  which  is  one  millionth  of  a  farad. 

(35)  C  (farads)  =^^). 

Ex.  12.  Determine  C,  Q,  and  E  for  the  following  given  values: 
(a)  C  =  .00015,  #=2000;  (b)  C  =  .0001,  #  =  110;  (c)  Q  =30,  #  =  550; 
(d)  C  =  .0002,  Q=  250. 

10.  The  capacity  C  of  a  condenser  varies  directly  with 

the  total  area  A  of  the  exposed  leaves,  and  with  the  dielectric 

constant  K,  i.e.,  the  specific  nature  of  the  insulating  material 

and  it  varies  inversely  with  the  thickness  d  of  the  dielectric. 

2248  . 
These  relations  are  expressed  in  (36)  in  which  is  a 

proportionality  factor: 

n     KA 

Ccc  — -. 

a 
2248KA 

m  ■      t-  Wd  . 

The  area  A  is  in  square  inches  and  may  be  expressed  as 
the  product  of  the  square  inch  area  of  one  leaf  times  the 
number  of  dielectric  separations,  d  expresses  the  dielectric 
thickness  in  mils  and  has  a  minimum  dimension  depending 
upon  the  potential  to  which  the  condenser  is  to  be  subjected 
in  practical  use.  A  standard  condenser  is  one  in  which 
the  air  serves  as  a  dielectric.     The  constant  K  for  air  equals 


514 


PRACTICAL  MATHEMATICS 


one  and  is  a  standard  for  reference  as  indicated  in  Table 
XXXVI. 

TABLE  XXXVL— DIELECTRIC   CONSTANTS 


Dilectric. 


Air  and  gases .... 
Manilla  paper 

Paraffin 

Beeswax 

Paraffin  oil 

Resin 

Ebonite 

India  rubber,  pure 

Turpentine 

Gutta  percha .... 

Shellac 

Sulphur 

Sperm  oil 

Olive  oil 

Glass 


Constant  k. 


1 

1.5 
1.68-2.30 

1.86 

1.92 
1.77-2.55 
2.05-3.15 
2.22-2.50 

2.23 
2.45-4.20 
2.74-3.60 
2.9  -4.0 

3 

3.1 
3.013-3.258 


Dielectric. 


Paraffined  paper .  . 

Mica 

Porcelain 

Quartz 

Castor  oil 

Tourmaline 

Crown  glass,  hard. 
Flint  glass,  light .  . 
Flint  glass,  dense  . 

Alcohol 

Methyl  alcohol .  .  . 

Glycerine 

Formic  acid 

Distilled  water .  .  . 


Constant  k. 


3.65 
4.00-8.00 

4.38 

4.55 

4.8 

6.05 

6.96 

6.57 
10.10 
26 
34 
56 
62 
76 


Ex.  13.  Prepare  a  table  for  the  above  dielectrics  which 
expresses  the  microfarad  (mf)  capacity  per  square  foot  of  area 
per  mil  thickness  of  dielectric. 

Ex.   14.     (a)  A  condenser  is  built  with  92  sheets  of  beeswaxed 

paper   7x15    ins.  and    2   mils   thick  which 

separate  the    alternately  connected  sheets 

of  tinfoil.     The  capacity  of  the  condenser 

is   1.47   microfarads.     Determine  the  value 

of  K.      (b)  Paraffined   paper   ll"xll"  is 

substituted  and  the   tin  foil  10"xl0"  is 

used  as  shown  in  Fig.  (211).     What  is  the 

capacity   of  the  condenser  when  100  sheets 

Fig.  211.— A  Leaf  of      are  used.      (c)  How  many   sheets  of    the 

Tinfoil  for  a  Con-      latter  must  be   used  to  give  a   capacity  of 

denser.  2  mf.? 

11.  The  resistance  of  a  condenser  is  measurable  and  is 
usually  expressed  in  megohms  per  microfarad  and  in  cable 
work  as  megohms  per  mile. 

Ex.  15.  In  Ex.  14  the  dielectric  resistance  is  160  megohms. 
Express  the  rating  of  the  condenser  in  megohms  per  microfarad. 


INDUCTANCE  AND  CAPACITY  515 

12.  Practical  Values  of  Capacities.  The  capacity  of 
3  miles  of  Atlantic  cable  is  about  1  mf.  The  capacity  of 
an  ordinary  overhead  wire  is  about  .03  mf.  A  pint  size 
of  Ley  den  jar  has  T^  mf.  and  a  quart  size  has  -j^-  mf. 
A  2-mf.  condenser  is  connected  with  an  office  telephone 
equipment.  Common  potentials  in  wireless  telegraphy  are 
30,000  volts  and  common  capacities  .014  mf. 

From  (35)  we  observe  that  the  quantity  of  electricity 
which  flows  into  a  given  condenser  is  proportional  to  the 
pressure  in  volts  which  is  applied  to  its  terminals.  If  the 
applied  pressure  is  supplied  by  an  A.C.  circuit  then  the 
rate  of  change  of  the  quantity  of  electricity  flowing  into  or 
out  of  the  condenser  is  proportional  to  the  rate  of  change 
of  the  E.M.F  impressed  at  its  terminals  as  expressed  in  (37). 

(38)  §-i. 

(39)  ,.     i  =  C% 

(39)  expresses  the  instantaneous  value  of  the  condenser 
current. 

The  quantity  of  electricity  which  flows  into  a  condenser 
of  capacity  C  farads  in  time  t,  when  the  impressed  E.M.F. 
E  is  constant  is  expressed  in  (40) : 


(40) 


=  CE\l-e  c«). 


(41)  q  =  CEe   ™. 

(41)  is  the  equation  for  the  dischagre  current  of  a  con- 
denser.    CR  is  called  the  time  constant  of  the  condenser. 

Ex.  16.  Construct  the  curves  of  charge  and  discharge  of  a 
condenser  of  2  microfarads  capacity  and  one  megohm  dielectric 
resistance  which  is  connected  to  a  110- volt  D.C.  circuit.  Plot 
these  curves  on  semilog  paper. 


516 


PRACTICAL  MATHEMATICS 


13.  Formulas  for  Calculating  Capacities.  The  micro- 
farad capacity  C  of  a  metal  sheathed  cable  is  expressed  by 
(42)  in  which  L>  and  d  are  the  respective  external  and 
internal  diameters  in  centimeters  of  a  dielectric  whose 
constant  is  K  and  whose  length  I  is  expressed  in  centimeters. 


(42) 


C  = 


2.4132ft 


107 logio 


D' 
d 


The  microfarad  capacity  C  of  an  aerial  line  of  twin  wires 
is  expressed  by  (43),  in  which  I  is  the  length  of  one  wire 
in  centimeters,  r  is  the  common  radius  of  the  conductors 
in  centimeters,  and  A  is  the  spacial  distance  in  centimeters 
between  the  conductors: 

(43)  C=       hm  A. 


10°  log 


10 


Ex.  17.  Transform  (42)  and  (43)  so  that  C  represents  the 
microfarad  capacity  per  1000  ft.  and  so  that  the  other  dimensions 
are  expressed  in  inches. 

14.  Parallel  Connection  of  Condensers.  A  variable 
condenser  may  be  constructed  by  mounting  one  set  of  plates 
on  a  movable  standard  so  that  the  exposed  areas  of  the 
plates  may  be  altered.  The  combined  capacity  of  con- 
densers, which  are  connected  in  parallel  equals  the  sum 
of  the  individual  capacities.  Suppose  the  three  con- 
densers shown  in  Fig.  212  have  capacities  Ci,  C2,  and  C3 


Fig.  212. — Condensers  Connected  in  Parallel. 

respectively,   and  corresponding  charges  Qi,   Q2,   and    Qz. 
The  potential  difference  E  is  uniform  for  each  condenser 


INDUCTANCE  AND  CAPACITY  517 

and  is  also  the  potential  difference  of  the  combined  con- 
densers whose  total  capacity  is  designated  by  C  and  whose 
total  charge  is  designated  by  Q.     Therefore 


(44) 

Qi  =  CiE, 

(45)     Q2  =  C2E, 

(46)     Qs  =  CzE, 

(47) 

Q  =  CE; 

(48) 

Q  =  Qi+Q2+Qs. 

(49) 

.*. 

CE  =  CiE+C2E+CzE. 

(50) 

.• 

.     C  =  Ci+C2+C3. 

(44)  .  .  .  (47)  are  derived  from  (35).  (48)  states  that 
the  total  quantity  of  electricity  passing  into  the  condensers 
equals  the  sum  of  the  quantities  passing  into  the  individual 
condensers.  The  interpretation  of  (50)  gives  the  law  for 
connecting  condensers  in  parallel;  the  total  capacity  equals 
the  sum  of  the  individual  capacities. 

15.  Series  Connection  of  Condensers.  If  three  con- 
densers Cij  C2,  and  C3  are  connected  in  series  they  will  all 
receive  an  equal  charge  Q  of  electricity,  but  their  potential 


Fig.  213. — Condensers  Connected  in  Series. 

differences  will  be  unequal  and  may  be  represented  by  E\, 
E2,  and  Ez  respectively.  C  is  the  total  capacity  and  E  the 
total  potential  difference,  therefore 

(51)     Q  =  CiEu  (52)  Q  =  C2E2,  (53)     Q  =  C3E3, 

(54)  Q  =  CE; 

(55)  E  =  Ei+E2+E3. 

ran  •    Q-Q+Q+Q 


518 


PRACTICAL  MATHEMATICS 


(55)  states  that  the  total  difference  of  potential  equals 
the  sum  of  the  differences  of  potential  across  the  individual 
condensers.  The  interpretation  of  (57)  gives  the  law  for 
connecting  condensers  in  series:  the  reciprocal  of  the  total 
capacity  equals  the  sum  of  the  reciprocals  of  the  individual 
capacities. 

16.  Condensers  Connected  in  Series-Parallel  Arrange- 
ment. The  arrangement  of  the  condensers,  shown  in 
Fig.  214,  consists  of  a  paralleled  pair  of  condensers  C\  and 


Fig.  214. — Parallel-Series  Connections  of  Condensers. 

C2  which  are  joined  in  series  with  a  third   condenser  C3. 
The  total  capacity  is  expressed  by  (58) : 


(58) 


1 


I 


1 


C     Ci+Ca^GY 

. C-i . 

c3 

1 1 

1 — 1 r 

Fig.  215. — Series-Parallel  Connections  of  Condensers. 


In  Fig.  215  C\  and  C2  are  connected  in  series  and  then 
placed  in  parallel  with  C\.  The  total  capacity  is  expressed 
by  (59): 

1 


(59) 


C  = 


1+JL 


■c2. 


Ex.  18.  The  diagram  (a)  .  .  .  (q)  in  Figs  2164  and  2165 
show  all  the  possible  arrangements  of  three  condensers  used  singly, 
doubly,  or  in  combination.  Determine  all  possible  values  for  three 
condensers  of  1,  2,  and  3  microfarads  respectively. 


INDUCTANCE  AND  CAPACITY 


519 


r& 

(a) 

r^n 

(m 

riz=-i 

(0 


id) 


1 


M 


(f) 


(0) 


(h) 


2 

3 

«      • 


Fig.  216A. — The  Possible  Arrangements  of  Three  Condensers. 


520 


PRACTICAL  MATHEMATICS 


(i) 


2 

-£EE 


w 


3 

2 E 


(j) 


(k) 


(m) 


(n) 


2 

3_         [-{= 


1  2 

-{^ 1= 


1  3 


(O) 


(P) 


2 

3 

1 

1 

1 

1 

1 

I 

1 

r^ 


(Q) 


Fig.  21QB. — The  Possible  Arrangements  of  Three  Condensers. 


INDUCTANCE  AND  CAPACITY 


521 


Ex.  19.  Determine  all  possible  capacities  which  may  be 
obtained  from  three  condensers  of  50,  100,  and  150  microfarads 
respectively. 

Ex.  20.  Determine  all  the  possible  values  of  capacity  which 
may  be  obtained  by  plugging  the  portable  adjustable  microfarad 
condenser,  shown  in  Fig.  217. 


c^: 


*fe 


#  1  #  #  # 


^ 


^3 


(),05  0.05  ,2  J  ,5, 

Fig.  217. — A  Portable  Adjustable  Condenser. 

Ex.  21.  Calculate  the  capacity  of  a  condenser  which  is  built 
with  sheets  of  tinfoil  4x5  ins.  and  mica  1  mm.  thick.  There 
are  200  mica  leaves  intervening  between  201  sheets  of  tinfoil. 


17.  Energy  Equations  of  an  A.C.  Circuit.  The  energy 
W  delivered  by  a  generator  to  an  A.C.  circuit  is  partly 
useful  power  represented  by  copper  losses  and  partly  wattless 
power  represented  by  the  energy  stored  in  the  magnetic 
and  electrostatic  fields.  The  useful  power  is  energy  which 
is  spent  in  the  heating  of  the  conductors  while  the  wattless 
energy  is  alternating  stored  and  recovered  from  the  field. 
The  useful  power  Wu  is  represented  by  the  PR  loss  in  (60). 
The  energy  WL  stored  in  the  magnetic  field  is  expressed  in 
(61)  and  the  energy  Wc  stored  in  the  electrostatic  field  is 
expressed  in  (62) : 


(60)       WU  =  PR.       (61)     WL  = 


LP 


(62) 


wc=c-¥ 


(63) 


W  =  PR+^+^. 


Interpret  (63),  (60),  (61)  and  (62)  in  which  W  is  the 
total  power  delivered  to  the  circuit  by  the  generator. 

18.  Reactance.  The  three  elements  E,  I,  and  R  enter 
in  the  discussion  of  a  D.C.  circuit  in  their  familiar  relation 


522  PRACTICAL  MATHEMATICS 

which  is  known  as  Ohm's  Law.  In  such  a  circuit  the  only 
continued  opposition  or  obstruction  offered  to  the  passage 
of  a  current  is  the  resistance  of  the  circuit.  In  an  A.C. 
circuit  the  opposition  or  obstruction  is  called  an  impedance 
(Z)  which  is  expressed  in  (64).  /  and  E  are  effective  values. 

(64)  /=§. 

Impedance  consists  of  three  distinct  elements,  viz.: 
resistance  R,  inductive  reactance  XL  due  to  inductance, 
and  capacity  reactance  Xc  due  to  capacity,  as  expressed 
in  (65).  Resistances,  reactances  and  impedances  are  measured 
in  ohms. 

(65)  z  =  VR2+(xL-xcy. 

As  in  a  D.C.  circuit  the  resistance  of  an  A.C.  circuit 
depends  upon  length,  cross-sectional  area,  and  specific 
resistance  of  the  conductor. 

Inductive  reactance  varies  directly  as  the  inductance  L 
of  the  circuit  and  the  frequency  /  of  the  current,  as  expressed 
in  (66).     The  proportionality  factor  is  2x. 

(66)  Xl  =  2tu/L  =  g)L. 

Capacity  reactance  varies  reciprocally  as  the  product  of 
the  frequency/  and  the  capacity  of  the  circuit  C,  as  expressed 

in  (67).     The  proportionality  factor  is 


(67)  Xc  = 


2% 
1 


2x/C     g>C 


(68)  Z  =  ^2+(WL-^)2. 

(69)  Z  =  ^+(2x/L-^): 


INDUCTANCE  AND  CAPACITY 


523 


(68)  and  (69)  result  from  the  substitution  of  (66),  (67)  and 
(65).     (70)  results  from  the  substitution  of  (68)  in  (64). 


(70) 


E 


4r2+{«l-M 


Observation.  In  general,  an  A.C.  circuit  is  subject  to 
change  due  to  resistance  and  reactances,  the  latter  caused  by 
inductance  and  capacity.  Inductive  reactance  has  a  positive 
sense,  whereas  capacity  reactance  has  a  negative  sense.  The 
total  reactance  of  a  circuit  is  the  excess  of  the  one  reactance 
over  the  other  and  bears  the  same  sign  as  the  greater  magnitude. 

The  total  reactance  of  a  circuit  is  designated  by  X,  as 
expressed  in  (71),  (72)  and  (73) : 

(71)  X  =  XL-Xe. 


(72) 
(73) 


E 


Vr2+x2' 

PR2+I2X2  =  PZ2=E2. 


Fig.  217a. 


CHAPTER  XXVI 
ELEMENTARY  OPERATIONS  OF  VECTOR  ANALYSIS 

1.  Vectors.  The  magnitude  of  a  quantity  may  be 
represented  graphically  by  a  line,  an  area,  or  a  volume. 
For  convenience  in  measurement,  magnitudes  are  indicated 
usually  by  the  lengths  of  straight  lines. 

When  a  suitable  length  is  chosen  for  a  unit,  each  magni- 
tude may  be  expressed  either  as  a  multiple  or  submultiple 
of  the  unit. 

With  the  aid  of  graphic  methods  it  becomes  possible  to 
compare  and  measure  areas,  masses,  densities,  temperatures, 
energy,  quantity  of  heat,  electric  charge,  potential,  rainfall, 
moonlight,  etc.  When  the  elements  under  consideration  are 
represented  by  any  arbitrarily  scaled  lines  they  are  called 
scalar  quantities.  A  number  of  quantities  of  the  same 
kind  may  be  united  or  combined  by  observing  the  funda- 
mental operations  of  algebra. 

There  are  many  other  elements  in  applied  science  which 
involve  direction  as  well  as  magnitude.  Thus  if  we  wish 
to  represent  motion  graphically,  it  is  not  sufficient  to  con- 
sider the  magnitude  of  the  displacement,  but  the  direction 
of  the  motion  must  be  indicated. 

Physical  quantities  that  involve  magnitude  and  direc- 
tion are  called  vector  quantities  or  simply  vectors.  Common 
illustrations  of  vectors  are  velocity,  accelerations,  forces, 
electric  currents,  magnetic  fluxes,  lines  of  force,  stress  and 
strains  in  structures,  flow  of  heat,  flow  of  fluids,  tides,  air 
currents,  etc.     Why  are  these  vector  quantities? 

524 


ELEMENTARY  OPERATIONS  OF  VECTOR  ANALYSIS   525 

In  specifying  a  force  it  is  necessary  to  know  its  magni- 
tude, its  direction,  and  its  sense,  which  involves  the  point 
of  application.  The  points  of  the  compass  serve  our  con- 
venience in  describing  direction.  The  bearing  of  a  line  is 
its  angle  of  deviation  from  a  north  and  south  reference  line. 

In  Fig.  218  the  line  c  bears  from  the  north  by  the  angle 
2  to  the  east.  The  horizontal  distance  of  the  extremity  of 
c  from  the  north  and  south  line  is  called  its  departure.  The 
vertical  distance  of  the  extremity  of  c  from  the  east  and 


r 

Departure  of  b 

1 

■ 

\1 

Departure  of  c 

°       ..2 

3 

V 

3 

Fig.  218. — The  Bearing,  Departure,  and  Latitude  of  a  Directed  Line. 

west  line  is  called  its  latitude.  When  the  bearing  and 
length  of  a  line  are  known  its  departure  and  latitude  may 
be  read  directly  from  a  traverse  table  or  they  may  be  com- 
puted by  means  of  the  respective  cosine  and  sine  of  the 
bearing  of  the  line. 

2.  Vector  Representation.  A  force  acting  vertically  is 
not  fully  described,  since  it  may  act  upward  or  downward. 
This  deficiency  in  orientation  is  obviated  by  marking  arrow- 
heads on  the  vectors  to  indicate  their  sense  as  shown  by  g 
and  h  in  Fig.  219.  A  vector  is  a  directed  line.  Its  length 
represents  the  magnitude  of  a  quantity  and  its  inclination 
or  bearing  to  an  arbitrary  line  of  reference  together  with 
its  arrow-head  shows  its  direction  and  sense  of  action. 

Fig.  219  represents  a  number  of  vectors  of  different 
magnitudes,  directions,  and  senses.  A  vector  diagram  may 
contain  vectors  which  represent  different  physical  quantities 
and  therefore  there  arises  a  necessity  for    distinguishing 


526 


PRACTICAL  MATHEMATICS 


them.  As  an  illustration  such  a  diagram  may  contain 
vectors  which  represent  E.M.Fs,  magnetic  fluxes,  and 
currents.  Vectors  may  be  distinguished  by  modifying  the 
arrow-head  and  the  feather,  as  shown  in  e,  f,  a,  t,  v,  r,  u. 
They  may  be  distinguished  by  drawing  them  with  light, 
medium,  heavy,  dotted,  and  double  lines,  as  shown  in  m, 
h,  n,  b,  and  c.  They  may  be  distinguished  by  marking  the 
lines  with  characters,  as  indicated  by  d,  q,  o,  s,  p. 

Vectors  are  designated  by  one  or  two  letters  placed  at 
or  between  their  extremities. 


Fig.  219.— Vectors. 


3.  Equal  Vectors.  A  vector  may  be  represented  arbi- 
trarily on  any  part  of  the  plane  of  our  paper  or  blackboard. 
We  assign  to  it  a  definite  length  for  its  magnitude  according 
to  a  convenient  scale  and  choose  an  arbitrary  line  of  refer- 
ence in  order  to  specify  its  direction.  A  vector  may  be 
represented  in  any  position  of  the  paper  by  moving  it  parallel 
to  itself  and  without  altering  its  magnitude.  Therefore 
vectors  are  equal  when  they  have  equal  magnitudes  and 
like  directions  and  the  same  sense.  Vectors  are  unequal 
when  one  of  these  conditions  is  altered.  Two  vectors  of 
equal  magnitude,  which  are  respectively  positive  and  nega- 
tive, are  parallel  and  of  equal  length,  but  their  arrow-heads 
will  point  in  opposite  directions. 

4.  The  Addition  of  Vectors.  The  two  extremities  of  a 
vector  may  be  distinguished  by  calling  them  the  head  and 
nock  or  beginning  and  end. 


ELEMENTARY  OPERATIONS  OF  VECTOR  ANALYSIS  527 

Two  or  more  vectors  are  added  graphically  by  placing 
them  consecutively  so  that  the  head  of  one  vector  is  in 
coincidence  with  the  nock  of  the  next  adjacent  vector.  A 
vector  which  joins  the  nock  of  the  last  vector  with  the  head 
of  the  first  vector  represents  their  sum.  The  vector  sum  is 
more  commonly  called  the  resultant  vector  or  resultant. 

The  resultant  of  two  or  more  vectors  is  a  vector  whose 
sense  opposes  that  of  the  continuous  sense  of  the  separate 
vectors  which  have  been  joined. 

In  the  left  of  Fig.  220  are  two  vectors  a  and  b  which  are 
are  to  be  added.     In  the  right  a'  and  b'  represent  a  and  b 


Fig.  220.— The  Addition  of  Vectors. 


respectively,  in  a  new  position  with  the  head  of  a'  coincident 
with  the  nock  of  b' .  R  is  the  resultant,  i.e.,  the  sum  of  a 
and  b  and  its  arrow-head  opposes  the  continuous  sense  of 
a!  and  V .  What  authority  enables  us  to  place  a  and  b  in 
their  new  positions  a'  and  b'  respectively? 

In  the  left  of  Fig.  221  are  two  vectors  a  and  b  which  are 
to  be  added.  The  addition  of  two  vectors  may  be  accom- 
plished irrespective  of  the  order  in  which  they  are  placed 
in  succession.  In  the  third  figure  a"  and  b"  represent  a 
and  b  respectively,  and  in  the  fourth  figure  a'"  and  b'" 
represent  a  and  b  respectively.  In  both  cases  the  resultant 
is  R,  showing  that  the  two  vectors  may  be  added  in  either 
order.  In  the  second  figure  the  resultant  R  is  drawn  as 
the  diagonal  of  a  parallelogram  which  has  been  constructed 


528 


PRACTICAL  MATHEMATICS 


upon  a'  and  bf  as  adjacent  sides.  Since  the  opposite  sides 
of  a  parallelogram  are  equal  it  will  be  seen  that  the  left 
and  right  halves  of  the  parallelogram  are  identified  with 
the  third  and  fourth  figures  respectively. 


Fig.  221. — Different  Arrangements  for  Producing  an  Equal  Resultant. 


Fig.  222  shows  five  different  methods  for  obtaining  the 
resultant  of  three  vectors  a,  b,  and  c.  The  three  vectors 
may  be  added  in  any  order  as  follows: 

In  (a)  the  sum  of  a  and  b  is  represented  by  Ri  and  the 
latter  is  added  to  c  producing  the  resultant  R. 

In  (b)  the  sum  of  a  and  c  is  represented  by  Ri  and  the 
latter  is  added  to  b  producing  the  resultant  R. 

In  (c)  the  sum  of  b  and  c  is  represented  by  Ri  and  the 
latter  is  added  to  a  producing  the  resultant  R. 

In  (d)  the  sum  is  obtained  in  one  operation  by  laying 
off  a,  b,  and  c  consecutively.  The  resultant  R  closes  the 
figure  and  completes  the  polygon. 

In  (e)  the  three  vectors  a,  b,  and  c  are  drawn  from  a 
common  point  G.  The  vectors  are  resolved  into  horizontal 
and  vertical  components  by  projecting  them  on  the  lines 
AK  and  GE  respectively,  i.e.,  a  departure  and  latitude  is 
determined  for  each  vector.  -\-KG  and  —  AG  are  the 
respective  horizontal  projections  of  c  and  6,  and  -\-FG  and 
-\-HG  are  the  respective  vertical  projections  of  b  and  c. 
a  is  vertical  and  has  only  a  vertical  projection.     The  alge- 


ELEMENTAKY  OPERATIONS  OF  VECTOR  ANALYSIS  529 


A  D  G  K 

Fig.  222.— The  Resultant  of  Three  Vectors. 


530  PRACTICAL  MATHEMATICS 

braic  signs  of  the  projections  follow  from  the  convention 
of  signs  in  trigonometry.  The  sum  of  the  vertical  pro- 
jections equals  EG  and  is  the  vertical  projection  of  the 
resultant  R.  The  sum  of  the  horizontal  projections  equals 
DG  and  is  the  horizontal  projection  of  the  resultant  R. 
Therefore  R  may  be  constructed  as  the  diagonal  of  a  paral- 
lelogram whose  two  adjacent  sides  equal  EG  and  DG  re- 
spectively. 

At  the  top  of  Fig.  223  are  five  vectors  A,  B,  C,  D,  and  E 
whose  resultant  R  is  obtained  in  (a),  (6),  (c),  (d)  and  (e) 
by  constructing  polygons  of  vectors  in  five  distinct  ways. 
In  (/)  the  two  vectors  A  and  B  were  added  and  their  re- 
sultant Ri  was  added  to  C,  producing  R2,  the  latter  was 
added  to  D,  producing  R3  and  R3  in  turn  was  added  to  E, 
producing  the  final  resultant  R.  (/)  is  the  method  of 
partial  or  continued  resultants.  In  (g)  the  final  resultant 
R  is  obtained  by  the  method  of  projections. 

Observation.  In  every  case  of  vector  addition  irrespective 
of  the  method  or  order  of  the  vectors  the  final  resultant  will  be 
the  same. 

Ex.  1.  By  the  polygon  method,  continued  resultant  and 
projection  method,  construct  the  resultant  of  the  following  groups 
of  vectors  illustrated  in  Fig.  223  :  (a)  A,  B,  C;  (b)  A,  B,  D;  (c) 
A,  B,  E;  (d)  B,  C,  D;  (e)  D,  C,  E;  (/)  C,  D,  E;  (g)  A,  C,  D;  (h) 
A,  C,  E;  (i)'  B,  D,  E.  Measure  and  compute  the  length  and 
bearing  of  the  resultant. 

Ex.  2.  Construct  the  resultant  of  A,  B,  C,  D,  and  E  in 
Fig.  223  by  five  polygons,  the  order  of  whose  sides  is  not  like 
those  shown  in  Fig.  223.  Obtain  the  resultant  by  the  continued 
resultant  method,  changing  the  order  shown  in  Fig.  223.  Measure 
the  length  and  bearing  of  the  resultant. 

Observation.  Two  vectors  which  are  added  constitute  two 
sides  of  a  triangle  and  the  resultant  the  third  side. 

Three  or  more  vectors  may  be  added  by  forming  a  polygon 
of  vectors  which  is  completed  or  closed  by  the  resultant. 

The  resultant  may  also  be  obtained  by  first  adding  two 
vectors  and  then  adding  another  vector  to  their  resultant.     This 


ELEMENTARY  OPERATIONS  OF  VECTOR  ANALYSIS  531 


Horizontal  Projection  of  R  =  Algebraic  sum  of 
horizontal  projections  of    A,  B,  C,  D,  E 


Fig.  223. — The  Resultant  of  Five  Vectors. 


532  PRACTICAL  MATHEMATICS 

process  is  continued  until  all  the  vectors  have  been  united. 
A  vector  which  places  several  vectors  in  equilibrium  is  the  nega- 
tive of  their  resultant. 

Another  method  for  combining  vectors  is  to  represent  them 
at  a  common  point  of  action  which  shall  represent  also  the  origin 
oj  two  rectangular  axes.  The  vectors  are  then  resolved  into 
horizontal  and  vertical  components  by  orthogonal  projection. 
The  algebraic  sum  of  all  the  horizontal  components  is  the  hori- 
zontal component  of  the  resultant.  The  algebraic  sum  of 
all  the  vertical  components  is  the  vertical  component  of  the 
resultant.  The  resultant  may  then  be  constructed  as  the  sum 
of  the  component  vectors. 

The  subtraction  of  vectors  is  identical  with  that  of  addition. 
The  vector  which  is  to  be  subtracted  has  its  sense  reversed  and 
then  as  in  algebra  it  becomes  a  case  of  addition. 

Ex.  3.  In  Fig.  224  Wh  W2,  and  Wz  are  weights  suspended  by 
strings  passing  over  pulleys  whose  friction  is  negligible.     Draw 


Fig.  224. — Forces  in  Equilibrium. 

the  vector  diagram  for  the  forces  showing  them  in  equilibrium: 
(a)  W1=5=W2,  W3=7;  {b)  Wx=3,  W*=5,  Ws=7;  (c)  ^  =  10, 
W2  =  15,  T73=20. 

5.  Concurrent  Forces.  Lines  of  vectors  having  a  com- 
mon point  are  concurrent. 

If  a  number  of  concurrent  forces  are  in  equilibrium  they 
form  a  closed  vector  polygon. 

A  closed  vector  polygon  indicates  that  the  forces  acting 
at  a  point  are  in  equilibrium. 

Ex.  4.  Forces  of  5  and  7.5  lbs.  respectively  act  at  a  point. 
What  is  the  resultant  when  the  angle  between  them  is  (a)  30°; 
(6)  60°;  (c)  90°;  (d)  120°;  (e)  160°;  (/)  180°? 


ELEMENTARY  OPEEATIONS  OF  VECTOR  ANALYSIS  533 

Ex.  5.  The  forces  acting  on  a  body  at  a  point  and  their 
respective  bearings  are:  a  =  20  lbs.  N.j  b  =  4  lbs.  N.E.;  c=13  lbs. 
30°  N.  of  E.;  d=17  lbs.  W.;  e  =  25  lbs.  S.E.  What  is  their  re- 
sultant and  what  force  acting  with  them  will  maintain  equilib- 
rium? 

Ex.  6.  A  string  is  fixed  to  a  point  A  and  passes  over  a  pulley  B 
2  ft.  distant  and  supports  a  weight  of  10  lbs.  A  ring  slides  along 
the  string  between  A  and  B  and  carries  a  weight  of  15  lbs.  Deter- 
mine the  tension  in  the  string  and  the  position  of  the  weight. 

Ex.  7.  A  street-car  with  its  load  weighs  10  tons.  The  average 
grade  of  the  street  is  5  per  cent.  Allowing  an  efficiency  for  the 
motors  and  gearing  of  75  per  cent  and  a  horizontal  speed  of  10 
miles  an  hour,  what  power  must  be  applied  to  the  motor?  See 
Fig.  225. 

>■  10  Mi.per  Hr.     I  =  l()0  O 

T 
W=  10  tons 

Fig.  225. — Forces  Acting  on  an  Inclined  Plane. 

6.  The  Vectorial  Representation  of  Currents  and  E.M.Fs. 
A  sine  curve  which  has  been  plotted  to  rectangular  coor- 
dinates may  be  replotted  upon  polar  coordinate  paper.  A 
sine  curve  which  has  been  plotted  to  rectangular  coordi- 
nates becomes  a  circle  passing  through  the  pole  in  the 
polar  representation,  as  shown  in  Fig.  226. 

In  Fig.  226  the  pole  is  located  at  O.  The  circles  whose 
diameters  are  OA,  OE,  OD,  and  OB  correspond  to  Eqs.  (1), 
(2),  (3),  and  (4)  respectively,  and  also  to  (5),  (6),  (7), 
and  (8)  respectively. 


(1) 

ei=.Esin  6. 

(5) 

i\  =  I  sin  9. 

(2) 

ei  =  E  cos  9. 

(6) 

12=1  COS  0. 

(3) 

63=  —  E  sin  8. 

(7) 

is=  —I  sin  0. 

(4) 

ei=  —E  cos  0. 

(8) 

U=  —  /  cos  0. 

534 


PRACTICAL  MATHEMATICS 


The  maximum  values  of  (1)  .  .  .  (8)  are  represented  by 
the  respective  diameters  OA,  OE,  OD,  and  OB.  The  in- 
stantaneous values  of  e\  .  .  .  e±,  %i  .  .  .  u  are  the  chords 
intercepted  in  the  respective  circles.  The  instantaneous 
values  of  e\  in  (1)  are  measured  by  chords  OE,  OF,  and 
OG  when  the  respective  angles  are  19°,  47°,  and  71°.     The 


Fig.  226.— The  Polar  Representation  of  Sine  Curves. 

angle  is  measured  between  the  chord  and  the  horizontal 
reference  line: 

when  0  =  19°,  then  ei  =  OE  =  E  sin  19°; 

when  6  =  47°,  then  ei  =  OF  =  E  sin  47°; 

when  e  =  71°,  then  ei  =  OG  =  E  sm71°. 

Each  chord  represents  an  instantaneous  value  of  the 
E.M.F.  in  (1)  and  is  a  vector,  since  it  has  both  magnitude 
and  direction. 

Observation.  The  ordinates  of  a  sine  curve  may  be  repre- 
sented by  the  chords  of  a  polar  circle  and  therefore  E.M.Fs. 
currents  and  magnetic  fluxes  may  be  represented  vectorily. 

OA,  OE,  OD,  and  OB  are  the  maximum  values  of  four 
sine  curves  whose  relative  displacement  is  90°.  If  the  four 
diameters  are  held  rigidly  together  and  rotated  about  O 


ELEMENTARY  OPERATIONS  OF  VECTOR  ANALYSIS    535 

their  respective  projections  are  the  simultaneous  values  of 
the  ordinates  of  the  four  corresponding  sine  curves.  The 
circles  may  be  dispensed  with  if  we  represent  the  maximum 
values  only  of  each  sine  curve  by  a  vector. 

The  angular  displacement  of  the  vectors  must  correspond 
to  the  phase  displacement  of  their  sine  curves.  The  effective 
values  of  E.M.F.  and  current  are  more  frequently  used  in 
calculation  than  their  maximum  values.  For  the  purpose 
of  graphic  calculation  in  A.C.  work  each  E.M.F.  and 
current  is  represented  by  a  vector  whose  length  equals  the 
magnitude  of  the  effective  values  of  the  respective  E.M.F. 
or  current.  A  diagram  containing  the  vector  representation 
of  E.M.Fs.  and  currents  is  named  a  clock  diagram. 

The  drop  across  a  resistor  is  measured  by  the  product 
of  its  resistance  times  the  current  flowing  through  it  and 
is  represented  by  a  vector. 

The  drop  across  a  reactor  is  measured  by  the  product 
of  its  reactance  times  the  current  flowing  through  it  and  is 
represented  by  a  vector. 

Since  reactance  drop  is  90°  out  of  phase  with  resistance 
drop  the  former  may  be  represented  by  a  vector  which  is 
drawn  at  right  angles  to  the  vector  which  represents  the 
latter  as  illustrated  in  Figs.  231  and  232. 


CHAPTER  XXVII 


VECTOR  ALGEBRA 


1.  Symbolic  Representation  of  a  Vector.  The  graphic 
representation  of  alternating  quantities  is  in  very  extensive 
practical  use.  Many  problems  may  be  solved  directly  by 
the  measurement  of  the  lines  of  a  vector  diagram.  There 
are  cases,  however,  where  the  measurements  of  lines  of  great 
disparity  in  magnitude  or  of  very  small  deviation  in  direc- 
tion would  cause  errors  not  permitted  in  technical  applica- 
tion. In  such  cases  the  numeric  values  of  the  vectors  are 
used  and  from  these  a  practical  result  is  obtained  in  numeric 

form.  We  shall  proceed  to 
examine  some  of  the  notations 
used  in  applying  algebra  to 
vector  diagrams. 

In  Chapter  XXVI  it  was 
shown    how  vectors    may  be 
resolved      into      components 
which  are  the  respective  pro- 
jections measured   along  and 
perpendicular    to    any    fixed 
axis.      Thus   in   Fig.  227  the 
vector  OV  has  a  magnitude  a 
and  it   is   resolved    into    the 
horizontal  component  a\   and  the  vertical   component  0,2. 
(1)  indicates    that  a  is  the  vector   sum  of  the  two  com- 
ponent vectors  a\  and  a^.     The  symbol  ©  means  vector  sum. 


Fig.  227. — Vector  Components. 


(i) 


a  =  ai©a2. 


536 


VECTOR  ALGEBRA  537 

2.  It  is  not  only  convenient  but  it  is  also  necessary  for 
calculations  that  we  should  be  able  to  distinguish  the  hori- 
zontal from  the  vertical  component.  In  order  to  provide 
for  this  necessity  Steinmetz  suggested  the  prefixing  of  the 
letter  j  to  the  term  representing  the  vertical  component, 
as  shown  in  (2) : 

Then 

(2)  V  =  ai+ja2, 

and  since  ai=acos(j>  and  a2  =  a  sin  4>,  then   (1)   and   (2) 
transform  into  (3)  and  (4)  respectively: 

(3)  a  — a  cos  <j> ©a  sin  <J>. 

(4)  a  =  a  cos  §-\-ja  sin  (J>. 

Expressions  (3)  and  (4)  indicate  that  the  vector  a  has  a 
component  a  cos  $  along  the  principal  axis  of  reference 
and  a  component  a  sin  $  measured  at  right  angles  to  the 
latter. 

Another  notation  which  has  been  suggested  by  Cramp 
and  Smith  is  to  distinguish  the  horizontal  component  from 
the  vertical  component  by  accents.  The  former  receives 
a  prime  or  single  accent  ( ' )  and  the  latter  a  double  prime 
or  second  accent  (  "  ).  Under  this  notation  (1)  and  (3) 
may  be  rewritten  as  (5)  and  (6)  respectively: 

(5)  a  =  a'+a". 

(6)  a  =  (a  cos  cj>) ' -f (a  sin  (j>) " . 

The  use  of  different  kinds  of  letters  for  the  two  com- 
ponents has  also  been  suggested  and  in  fact  a  number  of 
other  schemes  have  been  proposed,  but  instead  of  clarifying 
the  work  it  means  a  burdening  of  the  mind  and  a  difficult 
task  for  the  printer. 

3.  The  Magnitude  and  Inclination  of  a  Vector.  The 
vector  OV,  represented  in  Fig.  227,  has  a  magnitude  a 
and  an  inclination  cj>  with  the  axis  of  reference,  and  since 


538  PRACTICAL  MATHEMATICS 

a\  and  a^  are  the  lengths  of  the  component  vectors,  then 
the  length  a  of  OF  is  expressed  in  (7) : 

(7)  •  a  =  Va12+a22. 

The  inclination  §  is  determined  from  (8),  (9),  or  (10): 

(8)  tan  <[>  =  —. 

rr\\  •      jl       a2 

(9)  sin(!)=T 

(10)  cosc!>  =  ^. 

The  magnitude  of  the  resultant  c  of  two  vectors  of 
magnitude  a  and  b  respectively,  may  be  expressed  by  the 
law  of  the  oblique  triangle  in  (11)  and  (12),  where  §  is  the 
angle  between  the  vectors: 

(11)  c2  =  a2+62+2a&cosc[>. 


(12)  c  =  \/a2+b2+2ab  cos  4>. 

Ex.  1.  Express  the  value  of  the  impressed  E.M.F.  E  whose 
horizontal   component  is  ET  and  whose  vertical  component  is  Es. 

Ex.  2.  Express  the  value  of  the  current  /  in  a  circuit  whose 
horizontal  component  is  Ig   and  whose  vertical   component  is  Eb. 

Ex.  3.  Express  the  value  of  the  resultant  E  of  two  vectors 
Ei  and  E2  whose  phase  difference  is  60°. 

Ex.  4.  Express  the  value  of  the  resultant  E  of  two  vectors 
Ei  and  E2  whose  phase  difference  is  c[>.  Project  Ei  on  E2  and  also 
express  the  value  of  the  phase  angle  of  E  measured  from  E2. 


CHAPTER  XXVIII 
THE  GRAPHIC  SOLUTION  OF  A.C.  PROBLEMS 

1.  For  convenience  in  study  it  is  most  satisfactory  to 
separate  A.C.  problems  into  three  groups:  series  circuits, 
parallel  circuits,  and  series-parallel  circuits. 

In  general  A.C.  circuits  are  subject  to  change  due  to 
resistance  and  reactance.  They  are  therefore  influenced 
by  inductance,  capacity,  and  frequency. 

2.  Circuits  having  Resistance  and  Inductance.  A  counter 
E.M.F.  Es  due  to  inductive  reactance  Xl  may  be  expressed 
in(l): 

(1)  E/=-XLI=-2r,fLI. 

The  impressed  E.M.F.  E  must  possess  a  vertical  compo- 
nent Es  which  balances  or  overcomes  Es'  as  expressed  in  (2) : 

(2)  Es=-Es'  =  XlI  =  2tjLL 

The  impressed  E.M.F.  E  must  possess  a  horizontal 
component  Er  which  balances  or  overcomes  the  drop  across 
a  resistance  R.  In  this  sense  the  resistance  acts  as  a 
counter  E.M.F.  E'T  as  expressed  in  (3) : 

(3)  Er=-Er'. 

These  facts  are  illustrated  in  Fig.  228.  According  to 
KirchhofTs  Law:  The  resultant  of  all  the  E.M.Fs.  is  zero 
in  a  closed  circuit  if  the  counter  E.M.Fs.  of  resistance  and 
reactance  are  included. 

539 


540 


PRACTICAL  MATHEMATICS 


In  Fig.  228  OB  is  the  inductive  reactance  component 
of  OC  and  OA  is  the  corresponding  resistance  component. 
BC  and  AC  are  constructed  parallel  to  OA  and  OB  respect- 
ively : 
(4)  E=y/Er2+E,2. 

The  broken  lines  are  vectors  which  represent  counter 
E.M.Fs.     OB'  is  the  counter  E.M.F.  set  up  by  the  induct- 


Fig.  228— The  Balanced  E.M.F.  Vectors  for  a  Circuit  having 
Resistance  and  Inductive  Reactance. 


ance  and  OA'  is  the  counter  E.M.F.  due  to  resistance. 
OC  is  the  resultant  of  OA'  and  OC '. 

OB  balances  OB',  OA  balances  OA',  and  OC  balances  OC. 

Er  is  constructed  in  phase  with  the  current.  Since  the 
counter  E.M.F.  of  inductance  Es'  lags  90°  behind  the 
current,  then  for  counter  clockwise  rotation  Es'  will  be 
constructed  vertically  below  0.  Therefore  the  reactance 
component  Es  of  the  impressed  E.M.F  is  constructed  verti- 
cally above  0. 

Fig.  228  may  be  separated  into  two  parts;  one  of  these 
is  shown  in  Fig.  229.     The  impressed  E.M.F.  may  be  calcu- 


THE  GRAPHIC  SOLUTION  OF  A.C.  PROBLEMS    541 

lated  from  Er  and  Es  or  from  E/  and  Esf.  For  purposes 
of  simplicity  we  shall  choose  the  former  and  accordingly 
construct  Fig.  229  from  which  the  following  observations 
are  noted. 

The  impressed  E.M.F.  E  is  resolvable  into  two  compo- 
nents: one  of  these  represents  a  drop  of  potential  across  the 
resistance,  and  the  other  the  drop  across  the  reactance. 

To  overcome  the  resistance  R  an  E.M.F.  Er  =  IR  is  re- 
quired in  phase  with  the  current  represented  by  the  vector  OA. 

To  overcome  the  counter  E.M.F.  of  self-induction  an 
E.M.F.  ES  =  IXL  is  required  90°  ahead  of  the  current  repre- 
sented by  vector  OB. 


Fig.  229. — The  Components  of  an  Impressed  E.M.F.  in  an  Inductive- 
Resistive  Circuit. 

The  resistance  consumes  E.M.F.  in  phase  and  inductive 
reactance  an  E.M.F.  90°  ahead  of  the  current. 

The  current  I  being  in  phase  with  IR  may  be  represented 
upon  the  same  line  by  choosing  a  suitable  unit  for  R.  This 
is  shown  by  01  with  a  solid  head. 

The  current  I  lags  behind  the  impressed  E.M.F.  E  by  the 
angle  <&. 

3.  The  useful  power  P  which  is  consumed  in  a  circuit 
is  due  to  the  heating  of  the  resistance: 

(5)  P  =  PR  =  ^  =  ^  =  ErI  =  EI  cos  *. 


The  interpretation  of  (5)  states  that  the  active  power 
in  an  A.C.  circuit  may  be  computed  by  multiplying  the 


542  PEACTICAL  MATHEMATICS 

effective  values  of  the  E.M.F.  and  current  by  the  cosine 
of  their  phase  difference.  The  factor  cos  4>  is  called  the 
power  factor  of  the  circuit. 

4.  Circuits  having  Resistance  and  Capacity.  A  counter 
E.M.F.  Ecr  due  to  capacity  leads  the  current  by  90°  and 
therefore  an  impressed  E.M.F.  must  furnish  a  component 
Ec  which  balances  Ecr. 

(6)  &— &'. 

(7)  Ec^IXc  =  ^  =  -j^. 

(8)  /.     #,=  ---  =  __=-/Zc. 

In  Fig.  230  OB  is  the  capacity  reactance  component  of 
OC  and  OA  is  the  corresponding  resistance  component. 
BC  and  AC  are  constructed  parallel  to  OA  and  OB  respect- 
ively. 

(9)  E=VEr2+Ec2. 

The  broken  lines  are  vectors  which  represent  counter 
E.M.Fs.  OB'  is  the  counter  E.M.F.  set  up  by  the  capacity 
and  OA'  is  the  counter  E.M.F.  due  to  resistance.  OC  is 
the  resultant  of  OA'  and  OC. 

OB  balances  OB',  OA  balances  OA',  and  OC  balances  OC. 

Er  is  constructed  in  phase  with  the  current.  Since  Ec' 
leads  the  current  by  90°,  then  for  counter  clockwise  rotation 
Ec  will  be  constructed  vertically  above  0.  Therefore  the 
reactance  component  Ec  of  the  impressed  E.M.F.  is  con- 
structed vertically  below  0. 

It  is  customary  to  consider  the  impressed  E.M.F.  and 
its  components  only  and  therefore  the  lower  right-hand 
part  of  the  vector  diagram  is  used  for  simplicity. 

Observation.  E.M.Fs.  of  inductance  and  capacity  are 
opposite  in  phase,  i.e.,  180°  apart.  In  any  circuit  in  which 
they  are  both  present  their  effect  is  one  of  the   complete  or 


THE  GRAPHIC  SOLUTION  OF  A.C.  PEOBLEMS    543 


partial  neutralization.  Both  inductance  and  capacity  con- 
tribute to  make  the  total  reactance  and  therefore  we  may  dis- 
tinguish their  respective  reactances  by  subscripts. 

Es  =  El  =  component  of  E  which  overcomes  inductive 

reactance  voltage  =  IX L  =  2-zfLI 

Ec  =  component  of  E  which  evercomes  capacity 

reactance  voltage  =  —IXC=  ~Y7r' 


Fig.  230. — The  Balanced  E.M.F.  Vectors  for  a  Circuit  having  Resist- 
ance and  Capacity  Reactance. 

The  total  reactance  component  of  E  is  the  algebraic 
sum  of  the  inductance  reactance  voltage  and  the  capacity 
reactance  voltage. 

5.  Circuits  having  Inductance,  Capacity,  and  Resistance. 
The  vector  diagram  for  circuits  having  inductance,  capacity, 
and  resistance  is  a  combination  of  the  right-hand  halves  of 
Figs.  228  and  230,  as  shown  separately  in  Fig.  231.  There- 
fore the  vector  diagram  for  every  simple  circuit  will  contain 
the  following  vectors,  ET,  Es,  Ec,  and  their  resultant  E. 


544 


PRACTICAL  MATHEMATICS 


In  A.C.  vector  diagrams  an  open  arrow-head  indicates 
an  E.M.F.  and  a  closed  arrow-head  indicates  a  current. 

The  left-hand  diagram  of  Fig.  232  shows  an  excess  of 
inductive  reactance  and  a  corresponding  lagging  current. 
The  right-hand  diagram  of  Fig.  232  shows  an  excess  of 
capacity  reactance  and  a  corresponding  leading  current. 


\>\ 

— *i 

i 

i 

i 

x*X 

i 
i 

1 

i 

v 

Fig.  231. — The  Component  and  Resultant  E.M.F.  Vectors  of  a  Simple 

Circuit. 


Fig.  232. — Vector  Diagrams  for  Circuits  having  Resistance,  Induc- 
tance and  Capacity. 


6.  General  Expression  for  an  A.  C.  Circuit. 


(10) 
(11) 


_.  ,  T     impressed  E.M.F.  E 

The  current  I  =  — —. -, ~ 

impedance  Z 


1  = 


E 


Vr2+x2 


THE  GRAPHIC  SOLUTION  OF  A.C.  PROBLEMS    545 

E 


(12)  Z- 


4R2+[2%fL~M 


E 

(13)  -  i=Vr2+[xl+W 

(14)  I=i=EY' 

The  interpretation  of  (14)  states  that  J  equals  the  im- 
pressed E.M.F.  times  a  factor  Y  which  is  named  admittance 
and  is  defined  as  the  reciprocal  of  impedance. 


Ux/L-;     ] 


2x/L-    , 
'15)         cj)  =  tan-1| ^^ 


.X 

tan-1 


R 


Ex.  1.  What  are  the  magnitude  relations  for  the  inductive 
and  capacity  reactances  which  make  <}>  successively  positive,  zero, 
and  negative? 

Ex.  2.  Construct  a  right  triangle  showing  the  relation  between 
resistance,  reactance,  and  impedance,  and  also  <J>. 

Ex.  3.  Construct  a  right  triangle  similar  to  the  one  in  Ex.  2, 
substituting  admittance  for  impedance,  conductance  for  resist- 
ance, and  susceptance  for  reactance. 

Designate  conductance  by  g  and  susceptance  by  b. 

Ex.  4.  Express  cos  <1>  and  sin  $  in  terms  of  R,  X,  and  Z  from 
Ex.2. 

Ex.  5.  Express  g  in  terms  of  <J>  and  Z  and  substitute  for  Z 
cos  $  from  Ex.  4. 

Ex.  6.  Express  b  in  terms  of  <b  and  Z  and  substitute  for  Z 
sin  cj>  from  Ex.  4. 

Ex.  7.  Express  admittance  in  terms  of  g  and  b  from  the  relations 
in  Ex.  3. 

Ex.  8.  From  the  similar  triangles  in  Exs.  2  and  3  prove  the 
following: 

(16)      g=zi=WW>-  (17)      b=zi=lF+x>- 

(18)   H=i£»-      m   X=?=A>- 

Ex.  9.  Explain  why  diagrams  constructed  of  Z,  R,  and  X 
lines  or  y,  b,  and  g  lines  are  not  vector  diagrams. 


CHAPTER  XXIX 
SERIES  CIRCUITS 

1.  The  Symbols  for  R,  X,  Z.  Every  electrical  circuit 
possesses  resistance,  inductance,  and  capacity  and  there- 
fore reactance  and  impedance.  If  the  reactance  of  the 
circuit  is  negligible,  the  circuit  is  represented  with  resist- 
ance only  which  is  signified  by  using  the  symbol  R  in 
Fig.  233.     If  the  circuit  has  negligible  resistance  and  capacity 

reactance  this  fact  is  indicated  by  placing 
wwwwvwv  the  symbol  X,  Fig.  233,  in  the  circuit.  The 
^qq^q^qqqq^  symbol  X  is  also  used  when  only  the  capa- 
ooooofoonooQ        city  reactance  is  negligible.     If  the  circuit 

-n       ooo     ci     ,   i    nas  negligible  resistance  and  inductive  re- 
Fig.    233.— Symbols        ,  xt-,,.  ,.  ,  ,         .     . 

Used   in  an  A  C    actance  this  ±ac^  ls  indicated  by  placing 
Circuit.  the  symbol  C  for  capacity  in  the  circuit,  as 

shown  in  Figs.  233-235.  When  neither 
resistance,  inductive,  or  capacity  reactance  are  negligible  the 
three  symbols  for  R,  X,  and  C  may  be  used,  but  they  are 
merged  more  conveniently  into  the  impedance  symbol  Z 
Fig.  233.  The  use  of  these  symbols  is  intended  to  show 
that  the  distributed  resistances  and  reactances  of  a  circuit 
may  be  considered  as  concentrated  or  condensed  to  facilitate 
the  physical  interpretation  of  the  problem  without  causing 
appreciable  error  in  the  results.  The  power  factor  of  the 
circuit  is  abbreviated  P.F.  or  cos  §,  wherein  cj>  is  the  angle 
of  lead  or  lag. 

2.  Standard  Frequencies.  The  selection  of  a  suitable 
frequency  in  A.C.  work  depends  largely  upon  the  kind  of 

546 


SERIES  CIRCUITS  547 

load  to  which  the  power  is  furnished.  As  will  be  indicated 
later  every  circuit  has  a  definite  critical  frequency  at  which 
it  will  operate  most  efficiently.  It  is  customary  to  design 
A.C.  apparatus  to  meet  the  general  demands  of  practice 
for  which  60  and  25  cycles  have  been  adopted.  The  symbol 
for  cycles  is  written  (^). 

Ex.  1.    Determine  the  numeric  values  for  w  and  -   for  60 

~  and  25  ~. 

Ex.  2.  Determine  the  value  of  XL  for  a  60  ~  and  also  for  a 
25 ~  circuit,  Fig.  234,  when  L  equals:  (a)  .001/i;  (6)  .0015^;  (c) 
Mh;  (d)  ,015ft;  (e)  Ah;   (/)  .15/*. 


Fig.  234. — An  Inductive  Circuit. 

Ex.  3.  Determine  the  value  of  Xc  for  both  60  ~  and  25  ~ 
circuits  when  C  equals:  (a)  100  mf.;  (b)  50  mf.;  (c)  150  mf.;  (d) 
200  mf.;  (e)  250  mf.;  (/)  300  mf.  The  value  of  Xc  is  computed 
from  (1)  in  which  C  is  expressed  in  microfarads: 

105 
(1)  Xc=— . 

0)0 

101 
Ex.  4.     What  is  the  value  of  —  for  a  60^  and  25~  respectively? 

(0 

Ex.  5.     Construct  a  chart  of  ohmic  factors  whose  ordinates 

represent    w  =  2%f  and   whose   abscissas  represent  —  =  — — ;.     The 

w       2xf 

values  of  /  are  to  extend  from   15  ~  to  150  ~.     w  is  named  the 

106 
ohmic  factor  and  —  is  named  the  reciprocal  ohmic  factor. 

to 

Solve  the  following  problems  both  graphically  and  numerically 
for  60  ~  and  also  25  ~.  Determine  (a)  the  reactances  and  impe- 
dances of  the  entire  circuit  and  also  for  its  parts;  (b)  the  phase 
angles  and  power  factors  for  the  entire  circuit  and  also  for  its 
parts  indicating  lag  and  lead  by  +  and  —  angles  respectively; 
(c)  the  current  which  flows  through  the  circuit  when  110  voislts 


548  PRACTICAL  MATHEMATICS 

impressed  upon  it;  (d)  the  required  impressed  E.M.F.  to  pass 
10  amps,  through  the  circuit.  Arrange  the  diagram  of  the  circuit 
at  the  top  of  the  page,  showing  the  elements  of  the  circuit  and 
their  numeric  values,  as  given  in  the  data.  Show  the  graphic  and 
numeric  solutions  for  60  ~  and  25  ~  under  two  vertical  parallel 
columns. 

Ex.  6.  The  circuit  consists  of  a  non-inductive  resistance  of 
10 12.  The  vector  diagram  consists  of  a  current  line  OA  and  an 
E.M.F.  line  in  phase  with  OA.     Since  this  circuit  consists  of  resist- 


R  =>  10  ohms 


o< »T 


ance  only  there  will  be  no  phase  difference  between  current  and 
E.M.F.  Inductive  reactance  and  capacity  reactance  are  the  only 
factors  which  tend  to  cause  a  lag  or  lead  in  the  current : 

(a)      Z=R  =  10a.  (6)       $  =  0  for  both  60~  and  25~. 


~  -pi     1 1  n 

(c)  ^  =  v=~T7r  =  H  amPs-  when  110  volts  are  impressed. 

Z      10      = 

(d)  E=IZ  =  10X10  =  100  volts  required  to  produce  10  amps. 

Ex.  7.     The  circuit  consists  of  an  inductive  resistance  of  1012 
and  .01  henry.     Complete  the  omitted  calculations  and  check  all 


results  by  graphic  determination.     The  upper  diagrams  are  called 
impedance  diagrams  and  show  the  relations  between  R,  X,  and  Z. 


SERIES  CIRCUITS 


549 


R  =10  1=  10.3 

(i)    #  =  ioa 

(2)  XL  =  o>L  =  377  X. 01=  3.77a 

(3)  Z  -  V.R2+X2  =  VlOO+14.21. 
(a)     Z  =  10.68a 


(6)  cos^  =  r— 8  = 


(4) 

<l>  =         lag. 

(«) 

-     E      110      inQ 

/  =  Z  =  10.68=^3ampS 

(d)  E  =  IZ  =  10X  10.68  =  106.8  volts. 


R=10  1  = 

(1)  22  =  100. 

(2)  XL  =  *L  =  157  X. 01  =  1.57a 

(3)  Z  =  VlOO+2.47. 
(a)     Z  =  10.12a 


10.0 


(6)  cos  (J>  = 


10 


10.12 


(4) 


<{,= 


lHt£ 


/N  ,110  Mfl 

(c)  /  =  ^Q^2  ==  amps# 

(d)  #  =  10X10.12  =  101.2  volts. 


Ex.  8.  The  circuit  consists  of  a  non-inductive  coil  of  512  in 
series  with  an  inductive  coil  of  5fl  and  .01  h.  Show  that  the  dia- 
grams for  Ex.  8  are  identical  with  those  of  Ex.  7,  excepting  that 


IR  *  100 


the  resistance  drop  consists  of  two  equal  parts.     How  does  this 
diagram  enable  us  to  show  that  the  pressure  impressed  upon  the 


550 


PRACTICAL  MATHEMATICS 


circuit  is  not  equal  to  the  algebraic  sum  of  the  drops  between  the 
terminals  of  the  parts  of  the  circuit  but  is  equal  to  the  vector  sum 
of  the  separate  pressures. 

Observation.  For  every  series  circuit  there  is  an  impedance 
diagram  whose  three  sides  R,  X,  and  Z  are  in  the  same  ratio 
as  the  three  vectors  Er,  Ex,  and  E  of  the  vector  diagrams  for 
the  same  circuit. 


Ex.  9.  The  circuit  consists  of  a 
coil  which  has  .01  h.  inductance  but 
negligible  resistance.  Complete  the 
calculations  for  both  cycles  and 
explain  the  diagrams. 

60~ 

(1)  R=0. 

(2)  XL  =  377  X. 01  =3.77. 

(3)  Z  =  >/fi*+k*. 
(a)     Z=X=3.77SI. 


toL= 
3.77 


110  = 
IZ 


(6)  cos  §  =  -y  =  QL 


(4)        (J)  =  90  Mas;. 


lz= 

37.7 


1=29.18^ 


110 
3.77 


EX  =  IZ 

=  10X3.77  = 


R  =  0  L  =  01 


25^ 


1.57 


Eaf 
110  = 
IZ 


1=70.06 


EoT 
IZ  = 
15.7 


(1)  R  =  0. 

(2)  XL  =  1Z7. 
(a)     Z  =  1.45Q. 


Ex.  10.  The  circuit  contains  a  condenser 
of  100  mf .  capacity.  Complete  the  calculations 
and  explain  the  diagrams. 


C~100vif 


SERIES  CIRCUITS 


551 


z= 

26.53 


E2= 
110 


60~ 


(1)    R  =  0. 


__C8)X.-2|>--26.5 

(a)     Z  =  XC=  -26.512. 
0 


z  = 

63.68 


E«= 
265.3 


01=10 


(6)  cos  <J>  = 


(3) 


■26.5 


4>  =  90°  lead. 

r  =  H0  = 
26.5 

(d)  Ez  =  10X26.5  = 


(c) 


25~ 

(1)  R  =  0. 

(2)  Xc=^=M.7. 


E2= 
110 


1=  1.T3 


E== 


1=10 
0 


Ex.  11.     The  circuit  consists  of  a  resistance  of  10  fi  in  series 
with  a  condenser  of  100  mf .  capacity. 

Complete  the  calculations .  and  ex-  r  ^/V^M/WNA^^t 
plain  the  diagrams.  Check  the  nu- 
meric with  the  graphic  solution.  The 
upper  diagrams  are  called  impedance 
diagrams  and  indicate  the  magni- 
tude relations  between  R,  X,  and  Z. 


60~ 

(1)  ft  =  10  Q. 

(2)  Xc  =26.53  a. 

(3)  Z  =  Vft2+xc5 
(a)     Z  =28.35  n. 


(6)  cos4>=^ 


25' 


r      ] 

V*    ! 

x= 

\\\ 

23.53 

Y§> ! 

Nfei 

63.66 


R  =  10 


552  PRACTICAL  MATHEMATICS 

(4)      »- 

110 


\f     " 

Ex" 
266.3 

(c)       7  = 


28.35 


(d)  #  =  10x28.35 


Erioo 


Ex.  12.  The  circuit  consists  of  a  condenser  of  100  mf.  capacity 
in  series  with  an  inductance  of  .01  h.  The 
upper  diagrams  show  the  method  of  obtaining 
the  excess  of  reactances  which  is  a  negative 
or  capacity  excess  for  both  cycles.  The  solid 
arrow  head  indicates  current.  Complete  the 
calculations. 


27T/C 


26.53 


60~ 

WL=3.77 


=  22.75 


jfyt;LI  =  37J=EL 


1  =  10 


2W/CI~E< 

=  265.3 


E=E<fEs 

=  227 


(1)  XL  =377  x.01  =3.77  n. 

(2)  Xc=^= -26.53  a 

(3)  X=XL-XC  =  -22.75. 
(a)  Z=X  =22.75. 


(6)F.A'.=|  = 


0 


0. 


Z    X.l~Xc    = 

(4)     <t>=90°lead. 


25~ 

%0L  =1.57 


=  62.1 


SERIES  CIRCUITS 


553 


WL2 

7.53 


Ex.  13.  Perform  the  work  for  the  graphic  and  numeric  solution 
of  the  following  A.C.  circuits:  (a)  #=10 ft,  L  =  .015h.;  (b) 
#  =  15l2,L=.015h.;  (c)  #  =20 12,  L  =  .03  h.;  (d)  R  =3012,  L =  .03  h. 
(e)  #  =  1012,  (7  =  100  mf.;  (/)  #=1512,  (7  =  150  mf.;  (g)  #=2012 
(7=200  mf.;  (A)  #=3012,  (7=300  mf.;  (i)  #=5i2,  (7=50  mf. 
(j)  C  =  100  mf.,  L  =  .01  h.;  (A')  C  =200  mf.,  L  =  .02  h.;  (1)  C  =50  mf. 
L  =  .005h.  i 

Ex.  14.     Use  the  values  given  in  Ex.  13  for  a  frequency  of  100. 

Ex.  15.    Use  the  values  given  in  Ex.  13,  using  an  impressed 
voltage  of  120  volts. 

Ex.  16.  The  circuit  consists  of 
10  12  resistance  in  series  100  mf. 
capacity  and  .02  h.  inductance.  Con- 
struct the  impedance  diagram  shown 
in  shaded  lines.  Its  vertical  side 
represents  the  magnitude  X,  the 
excess  of  reactance,  its  horizontal 
side  represents  the  magnitude  of 
R  and  its  diagonal  side  represents 
the  magnitude  of  the  impedance  of 
the  circuit.  The  current  and  im- 
pressed E.M.F.  will  be  direction- 
ally  coincident  with  R  and  Z  re- 
spectively, and  their  phase  angle 
will  be  numerically  the  same  as  the 
^pgle  between  R  and  Z. 

Ex.   17.     The  circuit  consists  of  the  following  elements:    (a) 

#  =  1012,  (7=150  mf.,  L  =  .015  h.;  (6)  #  =  1012,  C  =  100  mf., 
L  =  .015h.;  (c)  #=1012,  C  =  100  mf.,  L=.03h.;  (d)  #=2012, 
(7  =  150  mf.,  L  =  .015h.;   (e)  #=2012,  (7  =  100  mf.,  L  =  .015  h.;    (/) 

#  =20 12,  C  =  100 mf.,  L  =  .03  h.;  (g)  R  =20 12,  C  =300 mf.,  L  =.03  h.; 
(h)  #  =  1012,(7  =  150  mf.,  L  =  .15h.;  (i)  #  =  1012,  (7  =  150  mf., 
L  =  .3h. 


wc 
526.5;: 


3.  Resonance.  A  series  circuit  is  said  to  be  resonant 
or  in  resonance  when  the  current  and  E.M.F.  in  the  supply 
circuit  are  in  phase.  This  means  that  the  current  attains 
its  greatest  value  and  the  impedance  formula  (2)  reduces 
to  (3).  It  also  implies  that  the  total  reactance  is  neutralized 
and  therefore  equals  zero  and  accordingly  (4)  follows  when 
X  =  0. 


554  PRACTICAL  MATHEMATICS 

E 


(2)  7  = 


VR2+X2 


(4)  2^-2^=°-      ;      . 

Solve  (4)  for  /.  The  resulting  equation  gives  the  critical 
frequency,  i.e.,  the  frequency  which  makes  the  circuit 
resonant.  The  natural  period  of  a  circuit  is  the  reciprocal 
of  its  critical  frequency. 

Ex.  18.  Determine  the  critical  frequency  and  the '  natural 
period  of  the  circuits  specified  in  Ex.  17. 

Ex.  19.  Construct  a  resonance  curve,  shown  in  Fig.  235, 
plotting  amperes  vertically  and  frequency  horizontally  for  a 
circuit  having  5  ohms  resistance,  an  inductance  of  .455  henry  and 
a  capacity  of  5.6  microfarads.    An  E.M.F.  of  100  volts  is  impressed 


-wwvw- 
r=5 


L=.455 


C=5.6m.f.< 


upon  this  circuit.  What  frequency  is  required  for  resonance? 
At  this  frequency  what  is  the  potential  difference  between  the 
terminals  of  the  condenser  and  that  across  the  inductance  coils? 
Ex.  20.  Construct  resonance  curves  for  the  circuits  described 
in  Ex.  17. 

4.  Problems  Solved  by  Adding  Impedances.  A  circuit 
may  have  any  number  of  pieces  of  apparatus  in  series 
each  of  which  may  or  may  not  possess  resistance,  inductance, 
and  capacity.  The  fundamental  principle  guiding  us  in 
the  investigation  of  such  a  circuit  is  that  when  an  E.M.F. 
is  impressed  there  is  but  one  current  which  has  the  same 
value  throughout  the  circuit.  The  pressure  at  the  terminals 
of  each  piece  of  apparatus  has  a  magnitude  and  phase 
dependent  upon  the  respective  values  of  Rs,  Ls,  and  Cs.  We 
have  seen  that  in  order  to  determine  the  pressure  necessary 


SERIES  CIRCUITS 


555 


to  send  a  definite  A.C.  through  such  a  series  we  must  add 
vectorially  the  pressures  required  to  send  the  current 
through  the  separate  pieces  of  the  circuit.  In  the  vector 
diagram  the  quantity  /  appears  as  a  common  factor  and 
therefore  the  problem  resolves  itself  into  the  addition  of 
impedances. 

Ex.    21.     The   circuit   consists   of   four   pieces   of   apparatus, 
#1=   80  ohms,  Li  =.2  henry,  C1=20  microfarads;  R2=50  ohms: 


2:>. 


18 


a 


it 


L=. 455  HENRY     R=5i2 

C  =  5. 6  MICROFARADS      E=110 

VOLTS       f  =  99.5  (CRITICAL) 

1 

> 

) 

< 

.     . 

' 

/< 2 
/  -- 

' 

1 

1 

kaSES 

1 

> 

°u 

)       '-< 

> 

10     *)     30     40     50 


70     80     90  -100   110    120  130   HO  150    100   170 
FREQUENCIES 


Fig.  235. — A  Resonance  Curve. 


Zv2=.25  henry;  C3=50  microfarads;  #4=75  ohms.  Determine 
the  impedance  for  the  four  respective  pieces  of  apparatus,  desig- 
nating them  by  Zt,  Zi,  Z3,  Z*.  Add  the  four  Zs  vectorially. 
Designate  the  terminal  pressure  of  the  four  parts  of  the  circuit 
by  Ei,  E2,  E3,  and  E*  respectively,  and  the  current  by  /.  The 
drop  across  any  impedance  is  expressed  in  the  general  formula  (5) 
from  which  (6),  (7),  (8),  and  (9)  follow: 


(5) 
(6) 


E  =1 X  impedance. 


Ei-i^+^-^-m 


556 
(7) 

(8) 

(9) 
(10) 
(11) 


PRACTICAL  MATHEMATICS 

#2  =/V#22  +  [<DL2]2  =IZ3. 


£'4=/V/W"2=/Z4. 

E^Z^Z&IZiQIZzQIZi. 

E=I(Zl@Z2®Z3®Zi). 


R,-80  R2=50 


1 

, 

»L,= 

wL2= 

94.25 

37.25    , 

31.4 
1    - 

£JLj= 

127.3 

75.4 

Ri 

=  80 

Ro= 

50   R 

(—  75 

— "' 1   v 

Z  = 

205.5 

1 
uCf 

132.6 

1        ' 
wC3~ 
53.06 
< 

1 
wC2~ 

318.3 

The  interpretation  of  (11)  states  that  the  current  times  the 
vector  sum  of  the  impedance  equals  the  impressed  E.M.F. 

The  combined  impedance  is  the  vector  sum  of  the  separate 
impedances  and  may  be  obtained  by  the  method  of  vertical  and 
horizontal  components,  i.e.,  by  Rs  and  Xs  respectively,  as  expressed 
in  (12)  and  (13): 


SERIES  CIRCUITS  557 

(12)  E  =/Z  =/\As  res)2 4  fs   indVctive  -S  capacity  l2. 

Vx^y  x-<      iU  v    v—  «*v    n  [—  reactances      ^reactances]  • 


(13) 


^  =  /^(^i+^2+/e4)2  +  L>(L1+L2)-V^+^l2. 


Substitute  the  data  from  Ex.  21  and  compute  E,  Eh  E2,  E3,  and 
Ei.     State  the  interpretation  of  (14) : 

(14)  Fn+E-i+Et+E^E. 

Ex.  22.     Complete  the  following  statements  from  observations 
upon  series  circuits: 

I.  Non-inductive  circuits  in  series. 

How  is  the  total  impressed  pressure  determined? 
How  is  the  total  resistance  determined? 
II.  Inductive  circuits  of  equal  time  constants  connected  in 
series. 
How  is  the  total  impressed  pressure  determined? 
How  is  the  total  impedance  determined? 

III.  Inductive  and  non-inductive  circuits  in  series  or  inductive 

circuits  of  unequal  time  constant  in  series. 

How  is  the  total  impressed  pressure  determined? 

How  is  the  total  impedance  determined? 

What  relation  exists  between  total  and  individual  pres- 
sures, also  between  the  total  and  individual  impedances? 

IV.  Condensers  connected  in  series  by  conductors  of  negligible 

resistance. 
How  is  the  total  impressed  pressure  determined? 
How  is  the  total  impedance  determined? 
V.  Condensers  in  series  with  non-inductive  resistances. 
How  is  the  total  impressed  pressure  determined? 
How  is  the  total  impedance  determined? 
What  magnitude  relations  exist  for  total  and  individual 
pressures,  also  the  relation  of  impedances? 
VI.  State  the  relations  for  condensers  in  series  with  inductances 

and  resistances. 
Ex.  23.     Make    complete  calculations    for  the  following  'cir- 
cuits  made   up   with  the  following  parts  of  Ex.  17:    (a)  and  (b) 
(a)  and  (c) ;    (a)  and  (d) ;    (a)  and  (e) ;    (a)  and  (/ ) ;  •  (a)  and  (g) 

(a)  and  (h);    (a)  and  (j);    (a)  and  (i);    (b)  and  (c);    (b)  and  (d) 

(b)  and  (c);    (b)  and  (/);    (b)  and  (g);    (b)  and  (h);    (b)  and  (i) 

(c)  and  (d);    (c)  and  (e);    (c)  and  (/);    (c)  and  (g);    (c)  and  (h) 
(c)  and  (i);    (d)  and  (e);    (d)  and  (/);    (d)  and  (g);    (d)  and  (h) 


558  PRACTICAL  MATHEMATICS 

(d)  and  (i);  (e)  and  (/);  (e)  and  (g);  (e)  and  (h);  (e)  and  (i); 
(/)  and  (gr);  (/)  and  (fc);  (/)  ard  (t)j  (fir)  and  (A);  to)  and  (i); 
(/&)  and  (i). 

Observation.  In  a  series  circuit  there  is  one  current  line 
which  is  constructed  horizontal  and  becomes  a  line  of 
reference  from  which  to  measure  the  phase  angles  of  the 
respective  E.M.Fs.  The  impedance  diagram  is  a  triangle 
similar  to  the  triangle  whose  sides  are  Er,  Ex,  and  E.  R's, 
X's,  and  Z's,  although  not  vectors,  are  added  vectorially.  By 
placing  arrow-heads  on  an  impedance  diagram  the  R,  X,  and  Z 
lines  become  respectively  Er,  Ex,  and  E.  The  ET  vector  repre- 
sents the  drop  across  resistance  R  and  is  always  in  phase 
with  the  current  I. 


CHAPTER  XXX 

PARALLEL  CIRCUITS 

1.  The  graphic  treatment  of  problems  relating  to  parallel 
circuits  is  analogous  to  the  treatment  for  series  circuits. 
In  the  preceding  chapter  series  circuits  were  solved  by 
adding  E.M.F.'s  vectorially  and  in  such  cases  the  current 
line  was  laid  off  as  the  horizontal  reference  line.  In  parallel 
circuits  the  E.M.F.  vector  is  the  horizontal  line  of  reference 
and  the  circuits  are  solved  by  adding  currents  vectorially. 
For  series  circuits  an  impedance  diagram  is  constructed, 
whereas  for  parallel  circuits  an  admittance  diagram  is  con- 
structed. 

2.  An  Equivalent  Impedance  is  a  single  impedance 
which  may  replace  several  impedances  in  a  circuit.  The 
equivalent  impedance  Z  of  a  group  of  parallel  circuits  whose 
individual  impedances  are  expressed  by  Zi,  Z2,  Z3,  and  Z4 
is  determined  indirectly  by  considering  their  respective 
admittances.  Admittance  has  been  denned  as  the  factor 
which  when  multiplied  into  the  terminal  voltage  of  a  circuit 
or  of  a  piece  of  apparatus  determines  the  current  passing 
through  it.  The  common  terminal  voltage  of  four  branches 
of  a  parallel  circuit  is  E  and  the  respective  currents  through 
each  circuit  is  expressed  in  (1),  (2),  (3),  and  (4): 

(1)  Zt-jUfFi. 

(2)  l2  =  ^  =  EY2. 

559 


560  PRACTICAL  MATHEMATICS 

(3)  I3  =  ZL  =  EY3. 

(4)  h  =  ~=EY4. 

The  total  current  /  flowing  through  the  mains  is  the 
vector  sum  of  the  currents  supplied  to  the  branches,  a 
expressed  in  (5) : 

(5)  7  =  710/20/30/4. 

(6)  /  =  £(Fi0F20F30F4). 

(7)  7=710720F30F4. 

(8)  /.     I  =  EY  =  ^. 

(6)  is  obtained  by  substituting  (1),  (2),  (3),  and  (4) 
in  (5).  The  interpretation  of  (6)  states  that  the  total 
current  /  may  be  obtained  by  multiplying  the  common 
terminal  voltage  by  F  the  vector  sum  of  the  admittances 
of  the  branches.  The  total  current  or  the  current  in  any 
branch  circuit  is  the  product  of  the  voltage  in  that  circuit 
times  its  admittance  as  expressed  in  (8). 

3.  If  we  consider  two  branch  circuits,  then  from  (1) 
and  (2),  we  have  (9),  which  states  that  the  total  current  / 
will  divide  vectorially  so  that  the  currents  I\  and  /2  will 
be  directly  proportional  to  the  admittances  of  the  respective 
circuits  or  inversely  proportional  to  their  impedances: 

The  phase  angle  of  each  circuit  may  be  obtained  from 
the  power-factor,  i.e.,  cos<i>  =  -~  The  phase  angles  are 
measured  from  the  horizontal,  E.M.F.   line.     In   Fig.  236, 


PARALLEL  CIRCUITS 


561 


Ii=EY\  and  I2  =  EY2  are  two  currents  whose  vector  sum 
is  I  =  EY.  A  parallelogram  having  its  two  sides  equal  to 
I\  and  h  respectively  has  a  diagonal  equal  to  I. 


Fig.  236. — The  Vector  Diagram  for  a  Divided  Circuit. 

Fig.  237  shows  the  current  /  resolved  into  a  component 
Ig  in  phase  with  the  E.M.F.  E  and  another  component  Ib 
perpendicular  to  the  E.M.F.  The  component  Ia  may  be 
obtained  by  multiplying  E  by  a  factor  g,  which  is  designated 
the  conductance  factor,  and  Ih  may  be  obtained  by  multi- 
plying E  by  a  factor  b,  which  is  designated  the  susceptance 
factor. 

Each  line  of  the  left-hand  triangle  of  Fig.  237  has  a 
common  factor  E  and  therefore  is  similar  to  the  admittance 


Ip-Egr  g     ^  y  -  conductance 

Fig.  237. — A  Current  Resolved  into  Two  Orthogonal  Components. 

triangle  which  is  the  heavily  shaded  figure  on  the  right. 
Therefore  (10),  (11),  (12),  and  (13)  follow  and  by  using 
subscripts  these  equations  apply  to  any  circuit  or  branch: 


(10) 


COS<}>=      y 


(12)  tan  <)>  =  -, 


(11) 
(13) 


sin  cj)  = 


V 


Y  =  Vg2+&. 


562  PRACTICAL  MATHEMATICS 

In  (10),  (11),  and  (12)  c|>  is  the  phase  angle  between 
current  and  E.M.F.  If  E  be  resolved  into  a  component 
Er  =  IR  in  phase  with  /  and  a  component  EX  =  IX  perpen- 
dicular to  Er,  then  the  three  vectors  will  constitute  a 
right  triangle  which  will  be  similar  to  an  impedance  triangle, 
containing  R,  X,  and  Z  and  cj>.  The  impedance  triangle 
and  the  admittance  triangle  are  similar  and  therefore  their 
homologous  sides  are  proportional,  as  expressed  in  (14) 
and  (15),  and  therefore  (16),  (17),  (18),  and  (19)  follow: 

(14)  |-i  (15)  I-*. 

(16)  ,      3  =  RY     R 


Z       Z2     R2+X2' 
XY     X  X 


(17) 

••       D       z       Z2     R2+X2' 

(18) 

.      r,     QZ      g          g 
"     a      Y      Y2     g2+b2' 

(19) 

bZ      b          b 
"             Y      Y2    g2+tf' 

(20) 

g~Y  co3$.            (21)          6=  7  sin*. 

(22) 

Y       a  /n2   1   h2  —        — 

1    v  *  ' b    z    vw+x* 

By  means  of  (16)  and  (17)  g  and  b  are  expressed  in 
terms  of  resistance,  reactance,  and  impedance.  By  means 
of  (18)  and  (19)  R  and  X  are  expressed  in  terms  of  con- 
ductance, susceptance,  and  admittance. 

In  the  following  problems  it  is  desired  to  have  the  following 
quantities  determined  for  each  branch,  as  well  as  for  the  total 
circuit  at  frequencies  of  60  ~  and  25  ~  :  (a)  The  impedance  and 
admittance;  (b)  the  current  which  flows  through  the  circuit  when 
120  volts  is  impressed  upon  the  circuit;  (c)  the  impressed  E.M.F. 
required  to  pass  10  amperes  through  the  mains;    (d)  the  phase 


PARALLEL  CIRCUITS 


563 


angle  and  power  factors  for  the  circuits.  The  circuits  are  dis- 
tinguished by  assigning  a  like  subscript  to  everything  connected 
in  the  same  branch. 

Ex.  1.  A  circuit  contains  two  non-inductive  resistances 
Ri=5il  and  #2  =  10  ft  in  parallel.  The  admittance  diagram 
consists  of  two  g  lines  which  represent  the  two  conductances  .2 
and  .1  respectively.    Their  sum  equals  .3,  and  therefore  the  two 


R.=  10  ohms 


circuits  may  be  replaced  by  a  single  circuit  with  an  equivalent 
impedance,  resistance,  and  reactance: 


(1) 


flfi  =y  =.2,         6i=0. 


(2) 

(3) 
(4) 

(5) 


gr2=^  =  .l,         62  =0. 

0=01 +02  =.3. 

Y  =  Vg2+b*=g=.3. 
Z4  =  l=3.33. 


Ex.  2.     A  circuit  consists  of  a  non-inductive  branch  of  10  n 
resistance  and  another  inductive   branch  of   .015  h.   inductance. 


Rj=  10 


nAA/vAAA/VV\AAAAAA* 


L.,     0.015 


Complete  the  calculations  and  explain  the  admittance  diagrams. 
Construct  the  vector  diagrams. 


564 


PRACTICAL  MATHEMATICS 


WL 
.424 


.177 


25  ~ 

(1) 

•-*■* 

V 

(2) 

&2=i-=.424. 
A2 

\ 

(3) 

Y  =  Vgi2+b2>=A3 

^\ 

(4) 

Z=y=2.32. 

y 

(5) 

I=EY=51.6. 

\ 

(6) 

E=IZ  =232. 

___J 

(7) 

COSc})=y  = 

BPQ> 

•■]    Si  -1 

(8) 

60  ~ 

V 

\ 

(1) 

01  = 

ji 

(2) 
(3) 

62  = 

r  =  W+&22  = 

Ex.  3.  The  circuit  consists  of  a  non-inductive  branch  of  512 
resistance,  and  a  second  branch  of  5S2  resistance  and  .015  h.  induct- 
ance.    The  impedance  and  angle  of  lag  may  be  determined  for 

Rr=5 

kMAAAA/WV\AAMAMM1 


R«=5  Lr^-015 


one  circuit  by  means  of  the  impedance  diagram.  The  lower  dia- 
gram shows  the  vector  sum  Y  of  the  admittances  Fx  and  Y2.  The 
angle  between  E  and  Y2  equals  the  angle  between  Z  and  R.  Con- 
struct the  vector  diagrams  and  explain.  Complete  the  calcula- 
tions. 


PARALLEL  CIRCUITS 


565 


R.f=5 


YpV 


(1)  <7i=7r=£; 

(2)  6x=0. 

(3)  g2 


— — — —  =  16 
(5)    0i +02=0  =  .36. 


(6)  r  =  V(gi+gO'+k'  =  .375. 

(7)  Z=2.7n. 

(8)  J=£F=4.5. 

(9)  £-=27. 


(10)    COSc}) 


i£_ 


(ii) 


*  = 


X=.2 


(1) 

*-*" 

(2) 

b2=0. 

(3) 

R* 

02  -  „   .     ,     v   . 

Ex.  4.     The  circuit  consists  of  two  inductive  branches  of  .01  h. 
and    .015  h.    inductance   respectively.     The   admittance   diagram 


R,=0 


L,=.oi 


OOMMOO^yMWOT^MK^I 


^&Qfl&&&&MMMM&&^^ 


R„=0 


would  consist  of  a  b  line  only  which  would  also  represent  the  y 
line.     Construct  the  vector  diagram. 


566 


PRACTICAL  MATHEMATICS 


Ex.  5.  The  circuit  consists  of  two  branches,  Ri  =  lOfi,  L\  =  .01  h., 
and  R2=5tt,  L2  =  .015  h.  respectively.  Determine  the  impedance 
of  each  branch  separately  and  combine  the  corresponding  recip- 


R,  =  10  Li=.01 


li^J^_0Q_0QOQQOQQQ_QQQQQ  J 

R„=5  U-.015 


rocals.  Calculate  6  and  g  for  each  branch  and  substitute  in  the 
formula  which  expresses  Y  in  terms  of  the  vertical  and  horizontal 
components  of  Yx  and  F2: 


Y  =  V(gi+g2)2+(bi+b2y. 

Ex.  6.  When  two  or  more  condensers  are  joined  in  parallel 
by  wires  whose  resistances  are  negligible  their  combined  effect  is 
equivalent  to  a  single  condenser  whose  capacity  is  the  sum  of 
the  capacities.  Illustrate  by  an  admittance  diagram  for  con- 
densers of  .01  and  .015  microfarad  respectively. 

Ex.  7.  The  circuit  consists  of  a  non-inductive  branch  Ri=l0 
ohms  and  a  capacity  branch  C2  =50  mf. 


R1=10 

A/WWVWWWV 


,WvWA 


25 


(1)  0i 

(2)  bx 

(3)  02 
(4)6, 


1. 


0. 
0. 

=  .00787. 


VW+622  =.10039. 
9.96a. 


(5)  Y  = 

(6)  Z== 

(7)  E=IZ  =  10X9.96  =9.96  volts. 

(8)  Z=ffF  =  12.05  amps. 

(9)  cos  cj>  = 

(10)  4>  = 


60  ~ 

(1) 

gi  =  ± 

(2) 

6x=0. 

(3) 

02=0. 

(4) 

62=.0188. 

PARALLEL  CIRCUITS 


567 


Ex.  8.  The  circuit  consists  of  two  branches,  #1  =5  12,  #2  =5 12, 
andC2=50mf. 

Ex.  9.  The  circuit  consists  of  two  reactive  branches  Ri  =  10 12, 
C,  =50  mf.,  and  #2  =15 12,  C2  =75  mf. 

Ex.  10.  The  following  branches  are  to  be  grouped  as  designated 
by  the  subscripts:  (a)  -1,  2;  (6)  -1,  3;  (c)  -1,  4;  (d)  -1,  5; 
(e)  -1,  6;  (/)  -1,  7;  (g)  -1,  8;  (A)  -1,  9;  (i)  -1,  10;  (j) 
-1,  11;  (fc)  -1,12;  (I)  -3,4;  (m)  -5,6;  (n)  -7,8;  (o)  -9, 
10;  (p)  -11,  12;  (?)  -2,  3;  (r)  -2,  4;  («)  -2,5;  (t)  -2,  6;  (w) 
-2,  7;  (t>)   -2,  8;  (w)  -2,  9;   (x)  -2,  10;  (y)  -2,  11;  (z)  -2,  12. 

(1)  #x  =  512,  L!=.01  h.;  (7)    #7=10l2,    L7  =  .005h.; 

(2)  #2  =  10l2,  C2=50  mf.;  (8)    #8  =  012,    C8  =  100mf.; 

(9)    #9  =  012,    L9-.015h.; 

(10)  #i0  =  10l2,  C10  =  100mf.; 


#x  =  512,  L!=.01  h.; 

#2  =  10l2,  C2=50  mf.; 

(3)  #3  =  512,  L3=.005h.; 

(4)  #4  =  1512,  C4  =  150mf.; 

(5)  #5  =  1012,  L5  =  .0105h.; 

(6)  #6  =  10l2,  C6  =  100mf.; 


(11)  #u=10l2,  Lu=.015h.; 

(12)  #12  =  012,  Ci2=150mf. 


If  a  number  of  circuits  are  placed  in  parallel  their  combined 
current  is  the  vector  sum  of  the  currents  in  each  branch.  The 
total  admittance  is  the  vector  sum  of  the  admittances  of  the  branches. 
The  total  admittance  is  expressed  in  (23)  and  is  obtained  by  adding 
the  square  of  the  sum  of  the  conductances,  i.e.,  the  g's  oft  the 
branches  to  the  square  of  the  sum  of  the  susceptances,  i.e.,  the 
6's  of  the  branches. 

(23)  F  =  VV+0*+0.+.  •  -)2  +  (bi+h+b3+.  .  .)*• 

(24)  i=EV(gi+g2+g3+gi)2+(bi+h+b3+biy. 

Ex.  11.  A  circuit  has  the  following  four  branches.  Determine 
the  current  in  each  branch  and  determine  the  total  current  by 
substituting  in  (24): 


Rr80o< 

»Cs* 

B*= 

9  R    -  ^ 

Lf  2  * 

>°  50 

50 

g    75   1                       ^ 

0,-20  2 

£  o 

cr 

§L4=|                       \L 

£l-2= 

1  :r 

sc<=< 

»    0    |     23 

P  o    <; 

#i=S0l2,  Lx=   .2h.,  Ci=20mf.; 
R2=  o,      L!=0,  C2=50mf.; 

#3=5012,  L3  =  .25h.,  C3=0; 
#4=7512,  L4=0,  C4=0. 


568  PRACTICAL  MATHEMATICS 

Ex.  12.  Combine  the  following  circuits  which  are  described 
in  Ex.  9:  (a)  -1,  2,  3,  4;  (6)  -5,  6,  7,  8;  (c)  -9,  10,  11,  12; 
(rf)  -1,  2,  3,  5;  (e)  -1,  2,  3,  6;  (/)  -1,  2,  3,  7;  (g)  -1,  2,  3,  8; 
(h)  -1,  2,  3,  9";  (t)  -1,  2,  3,  10;  (j)  -1,  2,  3,  11;  (&)  -1,  2,  3, 
12;  (I)  -1,  2,  4,  5;  (m)  -1,  2,  4,  6;  (n)  -1,  2,  4,  7;  (o)  -1, 
2,  4,  8;  (p)  -1,  2,  4,  9;  (g)  -1,  2,  4,  10;  (r)  -1,  2,  4,  11;  (*) 
-1,2,4,  12;  (0  -1,  2,  5,  6;  (u)  -1,  2,  5,  7;  (p)  -1,  2,  5,  8; 
(w)  -1,  2,  5,  9;  (a;)  -1,  2,  5,  10;  (?/)  -1,  2,  5,  11;  (2)  -1,  2,  5,  12. 

Ex.  13.  From  observations  upon  parallel  circuits,  state  the 
conclusions  for  determining  the  total  current  and  admittances  for 
the  following  conditions: 

I.  Non-inductive  resistances  in  parallel. 
II.  Inductive  circuits  of  equal  time  constants. 

III.  Non-inductive  and  inductive  resistances. 

IV.  Condensers  connected  with  negligible  resistance. 

V.  When   condensers   are   connected   in   parallel   with   non- 
inductive  resistances. 
VI.  When  condensers  are  in  parallel  with  inductive  circuits. 

4.  The  Circle  Diagram.  From  (12),  (14),  and  (15)  we 
derive  (24): 

(25)  tan<|>  =  ^  =  |. 

Formula  (25)  enables  us  to  compute  the  problems  for 
parallel  circuits  by  using  E.M.Fs  and  impedances.  It 
depends  upon  the  geometric  principle  that  an  angle  inscribed 
in  a  semicircle  is  a  right  angle.  Two  mutually  orthogonal 
sides  of  a  right  angle  are  used  to  determine  the  active  and 
wattless  components  of  the  total  current,  as  shown  in 
Fig.  238.  For  the  purpose  of  the  discussion  we  shall  use 
the  following  data:  Ri  =  1012,  Lx  =  .01  h.,  R3  =  512,  L3  =  .015  h. 

In  Fig.  238  OX  represents  the  impressed  E.M.F.  =  120 
volts  and  is  used  for  the  diameter  of  the  semicircle  OCDAX. 

From  O  lay  off   c^tan-1  -  =  tan-x  ^  =  20°  39'.     Then 
•  0i  Ri 

OA=I\R\  and  XA  =  ^L\I\.     The  current  in  branch  1  is  in 

OA 
phase  with  OA  and  is  represented  by  Ii=-^-.     From  O 


PARALLEL  CIRCUITS 


569 


lay  off  (^tan-1  -  =  tan-1  ^  =  48°  81'.     Then  OC  =  hRs 

and  XC^uLzIz.     The  current  in  branch  3  is  in  phase  with 

OC 
OC  and  is  represented  by  OH  =  Is  =  jz-.     The  vector  sum 

of  7i  and  Is  is  represented  by  I2  whose  phase  angle  cj>2  = 
36°  12'.     The   equivalent   resistance   drop   and   equivalent 


E=120 


Fig.  238. — The  Impedance  Solution  of  Parallel  Circuits, 
reactance  drop  for  the  combination  are  represented  by  OD 

and  DX  respectively.     The  equivalent  resistance  R  =  —f— 

12 

DX 

and  the  equivalent  resistance  =  —j—. 


Ex.  14.    Apply  the  principle  of  the  circle  diagram  to  the  so'u- 
tion  of  Exs.  8  and  9. 


CHAPTER  XXXI 

SERIES    PARALLEL    CIRCUITS 

1.  A  series  circuit  may  contain  a  number  of  groups  of 
impedances  in  which  each  group  consists  of  a  number  of 
impedances  in  parallel.  Such  a  circuit  is  known  as  a  series 
parallel  circuit.  The  equivalent  impedance  of  each  group 
is  determined  by  the  method  for  solving  a  parallel  circuit 
and  then  the  entire  circuit  with  its  equivalent  impedances 
is  treated  as  a  series  circuit.  When  several  impedances  of  a 
group  are  replaced  by  an  equivalent  impedance  Zn  the 
latter  has  an  equivalent  resistance  Rn  and  an  equivalent 
reactance  X„,  which  are  obtained  by  multiplying  the  im- 
pedance by  the  respective  cos  and  sin  of  the  phase  angle  §n 
of  the  group: 

(1)  Rn  =  Zn  cos  c[>n. 

(2)  X„  =  Znsin  <j*. 

Determine  all  the  data  and  properties  of  the  branches  and 
groups  as  well  as  for  the  entire  circuit  in  the  following  examples. 
Use  60~  and  25~  and  calculate  the  results: 

Ex.  1.  A  circuit  consists  of  an  impedance  B  in  series  with  a 
group  A.  The  group  A  consists  of  two  branch  circuits  Ri=50n, 
L\  =  50  milhenrys,  and  C2  =  50  mf .  The  impedance  B  consists  of 
an  inductance  L3  =  .015h.  The  following  equations  suggest  the 
method  of  procedure.     Check  by  vector  diagram : 

m  Rl  50  <S\     b  -    1    -  50  - 

W     9l    RS+XS    502+(18.85)2  Kf       2     -X2    2650 

(2)     6l=^^=502T(f85K  (4)   ^-vW(fc-fc)*. 

570 


SERIES  PARALLEL  CIRCUITS 


571 


(5)   ZA  = 


(12)    EA=ItZA. 


(6)  <{)A=cos-1— . 

Va 

(7)  Ra=Za  cos  4>a 

(8)  XA=£Asin4>A. 


(13)   EB=ItZB. 


(14)      7!=-^  lagging. 


#A 


(15)      h=w~  leading. 

Z-2 

~1R 

(9)  Ztotai  =  V(Ra  +Rb)2  +  (Xa  +Xb)*.    (16)  <!>i=cos     j -lag. 


(10)    ^tota^COS"1 


Ra+Rb 


'total 


(ID    /total  = 


7? 


(17)  ^=90°  lead. 

(18)  4>3=90°lag. 


Ri=50  Li-50 


Co-50 
R,=0  L,=.015 


Ex.  2.  The  circuit  consists  of  A  and  5  in  series.  A  consists 
of  two  branches  #i  =  10fl,  Li=.01  h.,  and  R2=oil,  L2  =  .015  h. 
B  consists  of  R,  =  1012,  L3  =  .005  h. 

Ex.  3.  The  circuit  consists  of  A  and  B  in  series.  A  consists 
of  two  branches  fti  =  15tt,  Li=.005  h.,  Ci  =  125  mf.,  and  fl2  =  10S2, 
L2  =  .01  h.,  C2  =  100  mf.  A  consists  of  tf3=5fl,  L3  =  .125h.,  C3  = 
100  mf. 

Ex.  4.  The  circuit  consists  of  A  and  B  in  series,  as  shown  in 
the  figure. 


Rj=15        Li-.005     Cj-US 


R.,-10  L4=,01  C4=  100 


572 


PKACTICAL  MATHEMATICS 


Ex.  5.     The  circuit  consists  of  A  and  B  in  series,  as  shown  in 
in  the  figure. 


Rj=15         Lj^.005      Cj=    125 


R2=10             L2=  .01  C2  =100 

H-j-0 L.w  0  C,-=100 


R4-=10         L4=.01l        C4  =  100 

Ex.  6.     The  circuit  consists  of  two  groups  A  and  B  in  series, 
as  shown  in  the  figure. 


R,=15         Li=.005      Cx=.125 


R4=0 


L4  =  .01        C4=0 


Ex.  ?.     In  Ex.  6  interchange  branches  2  and  4. 

Ex.  8.     In  Ex.  6  interchange  branches  2  and  3. 

Ex.  9.  In  Ex.  6  unite  branches  1,  2,  3,  into  group  A  and 
place  A  in  series  with  4. 

Ex.  10.  In  Ex.  6  unite  1,  2,  4  into  group  A  and  place  A  in 
series  with  3. 

Ex.  11.  In  Ex.  6  unite  1,  3,  4  into  group  A  and  place  A  in 
series  with  2. 

Ex.  12.  In  Ex.  6  unite  2,  3,  4  into  group  A  and  place  A  in 
series  with  1, 


CHAPTER   XXXII 
ALTERNATING   CURRENT   PROBLEMS 

Ex.  1.  Fig.  239  represents  the  cross-section  of  a  20-pole 
revolving  field  of  an  alternator.  How  many  revolutions  will  it 
require  to  give  frequencies  of  25~  and  60~  respectively? 

Ex.  2.  Ascertain  from  trade  catalogues  and  other  sources 
the  number  of  poles  adapted  for  a  line  of  commercial  alternators. 
From  the  information  compute  the  number  of  revolutions  which 
will  produce  25  and  60  cycles  per  second  respectively.  Tabulate 
the  data. 


Fig.  239.— Revolving  Field, 
of  Alternator. 


Fig.  240. — Inductor  Type 
Alternator. 


Ex.  3.  A  machine  of  the  inductor  type,  Fig.  240,  is  run  at 
1500  revolutions  per  minute  and  has  200  polar  projections  on  each 
end.     What  is  its  frequency? 

In  the  following  problems  make  a  sketch  of  each  circuit 
showing  the  symbols  for  the  different  apparatus.  Indicate  the 
numeric  values  of  the  supply  voltage  and  other  data.     Number 

573 


574 


PKACTICAL  MATHEMATICS 


each  formula  and  equation  and  arrange  the  work  neatly  and 
methodically  according  to  the  standard  set  for  this  department. 

Ex.  4.  Determine  the  power  factor,  P.F.,  for  each  of  the  follow- 
ing coils  which  are  connected  with  an  ammeter  to  110-volt  A.C. 
mains:  (a)  312  resistance  takes  15  amps;  (b)  2.5a  takes  12  amps.; 
(c)  1012  takes  10  amps.;  (d)  11012  takes  .2  amp.;  (e)  10012  takes 
1  amp.;   (/)  5012  takes  2  amps.;   (g)  25.312  takes  3.95  amps. 

Ex.  5.  Determine  the  resistance  of  the  following  coils  which 
are  connected  to  110-volt  A.C.  mains  and  have  the  following 
properties:  (a)  a  P.F.  =  .9 and  takes  25  amps.;  (6)  a  P.F.  =.2  takes 
10  amps. ;  (c)  a  P.F.  =  .85  and  takes  1  amp. ;  (d)  a  P.F.  =  .9  and  takes 
22.5  amps. 

Ex.  6.  What  current  will  pass  through  the  following  coils 
connected  to  110-volt  A.C.  mains:  (a)  a  P.F.  =1  and  resistance 
=  1012;  (6)  a  P.F.  =.9  and  resistance  5012;  (c)  a  P.F.  =.85  and 
resistance  =2512. 

Ex.  7.  An  ammeter  in  series  with  an  arc  lamp  on  an  A.C. 
circuit  indicates  6.3  amperes.  A  voltmeter  indicates  80  volts 
across  the  arc.  A  wattmeter  connected  to  the  lamp  circuit  shows 
450  watts  delivered  to  the  arc.  What  is  the  power  factor  of 
the  lamp? 

Ex.  8.    Two  coils  Fig.  241  are  joined  in  series  across  a  110 


P.F. -5 

1012  -R- 


P.  F.=.5 
R2-10i2 

Fig.  241.— Two  Coils  of  Equal  Time 
Constants  In  Series. 


A.C.  main.  Each  coil  has  a  power  factor  of  0.5  and  a  resistance 
of  10  ohms.     What  current  flows  through  the  coils? 

Ex.  9.  Determine  the  current  flowing  through  the  circuit 
and  its  power  factor  when  the  following  coils  described  in  Ex.  5 
are  connected  in  series:  (A)-(a)  (7>);  (B)-(a)  (c);(C)-(a)  (d); 
(D)-(b)  (c);  (E)-(b)  (d);   (F)-(c)  (d). 

Ex.  10.  Determine  the  total  current  flowing  into  the  circuit 
and  its  power  factor  when  the  following  coils  described  in  Ex.  4 
are  connected  in  parallel:    (A)-(a)  (6);    (B)-(a)  (c);    (C)-(a)  (d); 


ALTERNATING  CURRENT  PROBLEMS 


575 


(D)-(a)  (c);  (E)-(a)  (/);  (F)-(a)  to;  (£)-(&)  (c);  (H)-(b)  (d); 
(I)-(b)  (e);  (J)-(b)  (/);  (#)-(&)  to;  (L)-(c)  (d);  (M)-(c)  (e); 
(Ar)-(c)  (/);  (O)-(c)  (^);  (P)-(d)  (6);  (Q)-(d)  (/);  (fl)-(d)  to; 
OS)-(e)  CO;  (T)-(e)  to;  (£/)-(/)  to. 

Ex.  11.  The  current  coil  (C)  of  a  wattmeter  Fig.  242  has  a 
resistance  =  4. 5«  and  the  series  coil  (S),  has  a  resistance  =  1050ft. 
L  is  a  bank  of  lamps  each  of  which  takes  .5  amp.  when  the  supply 


Lamps 


Fig.  242. — Wattmeter  Connection. 


voltage  =  115.     (a)  How  many  watts  are  indicated  by  the  instru- 
ment when  1  lamp  is  burning;   (b)  when  two  lamps   are  burning? 
Ex.   12.     The  wattmeter  described  in  Ex.   11  is  reconnected 
as  shown  in  Fig.  243.     Determine  the  readings  of  the  instrument 


Fig.  243. — Wattmeter  Connection. 

when  (a)  one  lamp  is  connected;  (b)  when  two  lamps  are  connected. 

Ex.   13.     Two  alternators  A  and  B  are  connected  in  series. 

Each  has  an  E.M.F.  of  1000  volts.     The  two  machines  are  90° 

out  of  phase.    The  alternators  give  a  current  of  120  amperes 


A-1000V. 


I  =  120  A. 


1000  V. 


1000  V. 


120  A. 


B  =1000  V. 

Fig.  244. — Alternators  in  Series. 

which  lags  25°  in  phase  behind  the  resultant  E.M.F.  Determine 
the  resultant  voltage  and  the  output  of  each  generator.  Con- 
struct A  vertically  and  B  horizontally  as  the  adjacent  sides  of  a 


576 


PRACTICAL  MATHEMATICS 


parallelogram,  Fig.  244,  whose  diagonal  E  is  the  resultant  voltage . 
Then  construct  I  lagging  25°  behind  E. 

(1)  E  =  VA2+B*  =  1414  volts.  Data. 

A  =£  =  1000  volts. 

7  =  120  amps.     $=25°  lag. 

(2)  §a=70°,  <i>5=20°. 

(3)  Pa  =  Power  in  A  =EI  cos  70°  = 

(4)  PB  =  Power  in  B  =  EI  cos  20°.  = 

Ex.  14.  If  other  conditions  remain  the  same  as  in  Ex.  13, 
determine  (a)  the  required  voltage  in  the  machines  to  give  a  com- 
bined output  of  150  K.W.;  (b)  the  required  angle  of  lag  to  give 
a  combined  output  of  150  K.W.;  (c)  the  K.V.A.  of  the  machines; 
(d)  what  is  the  new  interpretation  of  power  in  each  machine  when 
A  is  135°  ahead  of  B? 

Ex.  15.  If  the  two  alternators  generate  voltages  of  1150 
and  1250  respectively,  then  their  resultant  voltage  will  be  the 
diagonal  (C)  of  a  parallelogram  whose  adjacent  sides  are  A  and 
B  as  shown  in  Fig.  245.  7  =  125  amps,  and  lags  behind  C  by 
an  angle  <[>i.  cj>i  and  4>3  are  the  respective  phase  angles  of  A  and 
B  with  C.  Determine  the  output  of  each  machine  when  (a) 
$  =30°;  (6)<|>  =25°;  (c)  cj>=20°;  (d)  when  A  and  B  are  135°  apart. 


no  v. 


C  =  85 


Fig.  245. 


Fig.  246. 


Ex.  16.  Two  coils  C  and  D  are  connected  in  series  across 
a  110-volt  A.C.  main.  The  drop  across  each  coil  is  85  volts. 
What  is  the  phase  difference  between  the  E.M.F.s.  of  C  and  D? 
The  supply  voltage  together  with  the  drops  across  C  and  D  form 
a  closed  vector  diagram  as  shown  in  Fig.  246,  which  is  at  the 
same  time  an  isosceles  triangle. 


a) 

(2) 


.    $    55 
Sln2-=85 


ALTERNATING  CURRENT  PROBLEMS 


577 


Ex.  17.  Two  coils  E  and  F  are  connected  across  a  110-volt  A.  C. 
main.  The  voltage  across  E  is  75  volts  and  the  voltage  across  F  is 
81  volts.     What  is  the  phase  difference  in  their  E.M.F.s.? 

Ex.  18.  Two  coils  G  and  H  are  connected  across  a  110-volt 
A.C.  main.  The  voltage  across  G  is  90  volts.  Their  phase  angle 
is  85°.     What  should  be  the  voltage  across  H? 

Ex.  19.  An  alternator  supplies  200  amperes  to  a  lamp  circuit 
and  delivers  80  amperes  to  start  an  induction  motor  whose  power 
factor  is  .3  at  starting.  What  is  the  total  power  delivered?  The 
lamps  and  motor  are  in  parallel,  as  shown  in  Fig.  240,  and  therefore 


150  A. 


P.F.-3 


Fig.  247. 

the  supply  voltage  #  =  110  is  laid  off  horizontally.  The  current 
II  in  the  lamps  =  150  amps,  and  is  in  phase  with  E,  whereas  the 
the  current  Im  in  the  motor  =  80  amps,  and  lags  behind  E  by  the 
angle  c{>  =  cos-x(.3).  The  total  current  I  is  the  vector  sum  of 
II  and  Im: 


(1) 


I=IL@Im. 


Resolve  /„,  into  a  horizontal  component  Img  and  a  vertical 
component  Imb  and  determine  I  from  the  sum  of  the  squares  of 
the  horizontal  and  vertical  components: 


(2) 

(3) 
(4) 


I  =  V  (150+80  cos  cj>) 2  +  (80  sin  c[>)2  = 


<kotal=COS 


^^O+SOcosJ) 


Aotal=^COS(j>total  = 

Ex.  20.  Determine  the  facts  described  in  Ex.  19,  making 
the  following  changes:  {a)  IL=200  amps.;  {b)  IL  =  100  amps.; 
(c)  7i=80  amps.;  {d)  Im=75  amps.;  (e)  Im=S5  amps.;  (/) 
P.F.  at  starting  =  .28;  (g)  P.F.  at  starting  =  .32;  (h)  P.F.  at 
starting  -  .25. 

Ex.  21.  Prepare  a  table  for  a  10-amp.  A.C,  giving  (a)  the 
rate  of  change  of  current  at  the  instant  it  passes  through  its  zero 
values;     (6)    the  value  which  the  current  would  attain  at  the  end 


578 


PRACTICAL  MATHEMATICS 


of  a  twelfth  of  its  cycle  if  its  zero  rate  continued  uniformly;  (c) 
its  actual  value  at  one-twelfth  of  the  cycle.  Calculate  for  grada- 
tions of  5  from  15~  to  60~  and  in  gradations  of  10  from  60~ 
to  150~. 

Ex.  22.  Prepare  a  table  for  currents  in  an  A.C.  circuit  supplied 
with  1 10  volts.  The  circuit  contains  a  constant  resistance  of  10ft 
and  a  variable  inductance  ranging  from  .01  h.  to  .1  h.  in  grada- 
tions of  .01  h.  (a)  Use  60—.  (b)  Use  25~.  (c)  Use  15-.  (d) 
Use  100~. 

The  tables  should  be  filled  out  under  the  following  headings, 
current,  inductance,  reactance,  power  factor,  phase  angle.  A  note 
of  the  constants  of  the  circuit  should  be  entered  above  the  table. 

Ex.  23.  A  circuit  has  a  power  factor  =  .9  and  is  supplied  with 
200  amps,  (effective)  from  110- volt  (effective)  A.C.  mains.  Deter- 
mine (a)  the  maximum  E.M.F.  Em,  (b)  the  maximum  current  Im, 
(c)  the  maximum  value  of  the  power  P+m,  and  (d)  the  minimum 
value  of  the  power  P~m: 


(1) 


F    -E   - 
Em    .707 


(2) 


/»  = 


.70, 


In  Fig.  248  the  vectors  represent  effective  values,  whereas  in 
Fig.  249  the  vectors  represent  maximum  values.    The  vertical 

E=100J 


110  V 


200  A 


P.F.— .9 

Fig.  248. — The  Usual  Representation 
with  Effective  #  and  7. 


Fig.  249. — Maximum 
E.M.F.  and  Current. 


dotted  lines  in  Fig.  249  are  the  instantaneous  values  e  and  i  of 
E.M.F.  and  current  respectively: 


(1) 
(2) 


P+m  =EmIm  COS*  ~2  =EI(1  +P.F.)  = 

P_ro  =  -EmIm  sin'  %  =  -EI(1  -P.F.)  = 


Ex.  24.  Determine  the  maximum  and  minimum  power  of  the 
following  alternators:  (a)  E  =220,  I  =500,  P.F.  =.8;  (6)  #  =  110, 
7=500,  P.F.=.8;  (c)  #=500,  7=500,  P.F.  =.8;  (d)  E  =220, 
7=500,  P.F.=.75;  (e)  #  =  1000,  7=500,  P.F.=.8;  (/)  #  =  110, 
7=50,  P.F.  =.95. 


ALTERNATING  CURRENT  PROBLEMS 


579 


Ex.  25.  Determine  the  power  factor  of  the  circuits  which  are 
supplied  by  alternators  as  follows:  (n)  P+m=50  K.W.,  P-m  = 
-5  K.W.,  Em  =  120  volts;  (b)  P+m  =  55  K.W.,  P-m  =  -7.5  K.W., 
#=220  volts;  (c)  P+m=50  K.W.,  P_m=-5  K.W.,  Im=  350 
amps;  (d)  P+ro  =  55  K.W.,  P-m  =  -7.5  K.W.,  /  =  225  amps. 


C=-  .00001 


Fig.  250. 

Ex.  26.  Determine  the  active  and  wattless  component  of  the 
power  supplied  to  a  fan  motor  which  is  connected  in  series  with 
a    non-inductive     rheostat    across     110- volt    mains.  x 

When  the  fan  motor  takes  1  amp.  the  drop  across  it 
is  64  volts  and  the  drop  across  the  rheostat  is  625. 
volts.     Explain  the  vector  diagram  in  Fig.  250. 

Ex.  27.    An   A.C.  has   a  maximum  value  =  135.5 
amps,  and  is  connected  to  a  series  circuit  having  a 
resistance  R,  an  inductance  L,  and   a   capacity    C. 
Draw  the  impedance   and  vector  diagrams  for  60~ 
and    25~    and   determine   the    phase   and    effective 
values   of   the  E.M.F.  across  Rh  XL,  and  Xc  for  the 
following:     (a)  #  =  10fi,  L  =  .001  h.,  C  =  . 00001  F;    (b) 
R=5Q,  L  =  .001h.,  C  =  10 
mf.;    (c)  fl=250,  L  =  .001 
h.,  (7  =  10   mf.;     (d)  R  = 
2.5fi,  L  =  .01    h.,    C  =  100 
mf.;  (e)  #  =  10fl,    L  =  .025 
h.,  C  =  10mf.  (/)  R  =  10n, 
L  =  .02h.,  C  =  10mf.;  (g)  R  =  10n,  L  =  .005h.,  CI  =0  mf.    Deter- 
mine the  critical  frequency  for  each  circuit  and  also  determine  the 


580 


PRACTICAL  MATHEMATICS 


required  reactance  which  should  be  added  to  each  circuit  to  make 
it  resonant  at  25~. 

Ex.  28A.  An  A.C.  circuit  is  supplied  with  10  amps,  at  110  volts. 
The  current  leads  the  E.M.F.  by  25°.  Determine  the  resistance, 
reactance,  and  impedance  of  the  circuit  graphically,  and  numer- 
ically : 


(1) 


E 


(2) 
(3) 


fl=Zcos4>=Zcos  25°  = 

X  =--Z  sin  <j>  =  Z  cos  25°  =  R  tan  25* 


In  Fig.  251  lay  off  R  horizontally  and  Z  25°  behind  R,  then 
construct  XC1R.  The  scale  R,  X,  Z  is  determined  from  (1), 
(2),  or  (3). 


Fig.  251. 


Ex.  28B.  Explain  why  Fig.  252  represents  the  impedance 
diagram  for  the  circuit  described  in  Ex.  25,  when  the  current 
lags  25°  behind  the  E.M.F.  How  can  the  impedance  diagram  be 
supplemented  in  order  to  become  the  vector  diagram  of  the  circuit? 

Ex.  29.  Determine  the  resistance,  reactance,  and  impedance 
of  A.C.  circuits  which  are  supplied  with  15  amperes  at  110  volts 
when  the  phase  angles  are  as  follows:  (a)  30°:  (6)  35°;  (c)  45°; 
(d)  55°;  (e)  75°;  (/)  -15°;  (g)  -5°;  (h)  -35°;  (i)  -45°; 
(?)  -55°;   (k)  -75°;   (I)  15°;   (m)  5°;   (n)  90°;   (o)  0°. 

Determine  the  capacity  of  the  above  circuits :  when  (I)  the 
inductance  is  .01h.  and  the  frequency  =60~;  (II)  when  the 
capacity  is  100  mf.  and  the  frequency  =  25~. 

Ex.  30.  Determine  the  resistance,  reactance,  and  impedance 
of  A.C.  circuits  which  are  supplied  with  10  amps,  at  110  volts 
when  the  power  factors  are  as  follows:  (a)  .7;  (6)  .85;  (c)  .8; 
(d)  .9;  (e)  .95;  (/)  .75;  (g)  .55;  (h)  .96;  (i)  .97;  (j)  .98;  (k)  .985; 
(m)  .99;  (n)  .995;  (o)  1.  Determine  the  capacity  of  the  above 
circuits  when  the  inductance  =  .02h.  and  the  frequency  =25~. 


ALTERNATING  CURRENT  PROBLEMS 


581 


Ex.  31A.  A  lamp  circuit  Fig.  253  receives  100  amperes  from 
an  A.C.  main.  An  inductive  circuit  of  negligible  resistance  takes 
15  amperes  when  connected  across  the  mains. 
What  is  the  total  current  delivered  to  the 
mains?  h  is  in  phase  with  the  E.M.F. 
whereas  h  is  normal  to  the  E.M.F.  and 
therefore  I  is  expressed  by  (1).  Determine  the  power  factor  of 
the  circuit. 


(1) 

7  =  V/i2+/2 

6 

u  66666606 

3  Ii=15 

I2=100 

Vl002  +  152 


I2=15 


i9=ioo 


Fig.  253. 


Ex.  31B.  Determine  the  power  factor  and  the  total  current 
supplied  to  the  circuit  shown  in  Fig.  246  when  the  lamp  current 
1 2  has  the  following  values:  (a)  50  amps.;  (b)  15  amps.;  (v)  75 
amps.;  (d)  60  amps.  Calculate  the  admittance,  susceptance,  and 
conductance  of  the  circuit.    Why  is  Y  shown  behind  g  in  Fig.  253? 

Ex.  31C.  Determine  the  power  factor  and  the  total  current 
supplied  to  the  circuit  shown  in  Fig.  253  when  the  inductive 
current  7\  has  the  following  values:  (a)  10  amps.;  (b)  20  amps.; 
(c)  25  amps.;  (d)  5  amps.  Calculate  the  admittance,  susceptance, 
and  conductance  of  the  circuit. 

Ex.  32A.  A  capacity  of  60  mf .  and  an  inductance  of  .025h.  are 
placed  in  series  with  a  resistance  of  10ft.  (a)  Determine  the 
frequency  which  will  make  the  circuit  resonant.  (6)  At  the 
critical  frequency  what  is  the  drop  across  the  condenser  when 
a  110-volt  A.C.  is  impressed  upon  the  circuit? 

Ex.  32B.  If  the  capacity  is  halved  and  the  inductance  doubled 
in  Ex.  31A  what  is  the  effect  upon  the  calculated  values  (a)  and  (6)? 

Ex.  33.  Determine  the  admittance,  conductance,  and  suscep- 
tance of  a  circuit,  one  branch  of  which  con- 
tains .25  h.  and  5ft  whereas  the  other  branch 
contains  50  mf.  and  1ft.  Fig.  254  represents 
the  admittance  diagram  when  the  frequency 
=  25~.  When  the  capacity  reactance  equals 
the  inductive  reactance  of  the  branches  the  parallel  circuit  is 
said  to  be  resonant. 


582 


PEACTICAL  MATHEMATICS 


3.0 


:.'.« 


&« 


2.4 


:>.:> 


2.0 


CO 
Q. 

z 

UJ 

QC 

D  1.2 
O 


10 


0.8 


(j.i 


0.2 


E=110 

VOLTS 

X=37. 

r  OHMS 

/=« 

I  CYCLE 

5 

10  20  30  40  5U  00  70 

RESISTANCE  IN  OHMS 


80  90  100 


Fig.  255. — An  Inductive  Circuit  with  Variable  Resistance. 


ALTERNATING  CURRENT  PROBLEMS     583 

Ex.  34.  Make  the  graphic  and  numeric  calculations  for  the 
circuit  described  in  Ex.  33  when  (I)  the  following  values  are  sub- 
stituted for  the  inductance:  (a)  .2h.;  (6)  .3h.;  (II)  when  the 
following  values  are  substituted  for  the  capacity:  (c)  25  mf.;  (d) 
100  mf. 

Ex.  35.  Calculate,  tabulate,  and  plot  a  curve  as  shown  in 
Fig.  255  between  current  and  resistance  in  an  inductive  circuit. 
The  circuit  is  supplied  from  a  60-cycle  110- volt  main.  The 
resistance  ranges  from  0  to  10012.  and  the  inductance  has  the 
following  constant  values :  (a)  L  =  .1  h. ;  (b)  L  =  .3  h. ;  (c)  L  =  .5  h. ; 
(d)  L  =  .75  h.  (e)  L  ='lh.;  (/)  use  the  value  of  L  given  in  (a)  but 
plot  R  between  0  and  1000a 

Ex.  36.  A  transmission  line  delivers  100  amperes  to  a  non- 
inductive  circuit.  The  E.M.F.  across  the  terminals  of  the  latter 
is  10,000  volts.  The  transmission  line  has  a  resistance  =  512  and 
a  reactance  =  2.512.  (a)  What  is  the  generator  voltage?  (b)  What 
is  the  phase  difference  between  generator  E.M.F.  and  receiver 
voltage? 

Ex.  37.  Make  the  following  substitutions  in  Ex.  36  and 
determine  (a)  and  (b)  when:  (I)  line  resistance  =4i2;  (II)  when 
line  resistance  =3.512;  (III)  when  line  reactance  =212;  (IV)  when 
line  reactance  =2.2512;  (V)  when  line  reactance  =  1 12;  (VI)  when 
line  reactance  =512. 

Ex.  38.  A  transmission  line  with  a  negligible  inductance  has  a 
resistance  =  512  and  delivers    100  c 

amps,  to  a  receiver  circuit.  The 
latter  is  entirely  inductive  and 
has  10,000  volts  at  its  terminals. 

In  Fig.  256   EC=BC  is   the   re-  RL=5ohmS    iooAmps. 

ceiver  voltage,  El=AB  is  the 
line  drop,  and  EC=AC  is  the 
generator   voltage  and   therefore  Fig.  256. 

(1)  follows: 


(i)  eg=\/Ec2+el>  = 


u 


Ex.  39.    What  would  be  the  effect  upon  the  genera-  Ei 

tor  voltage  if  the  inductive  resistance  of  the  receiving  circuit  in 
Ex.  38  were  replaced  by  an  equal  capacity  reactance. 

Ex.  40.  A  transmission  line,  Fig.  257,  with  negligible  induct- 
ance has  a  resistance  of  4.512  and  delivers  100  amps,  at  10,000  volts 
to  a  receiving  circuit  whose  power  factor  =  .7.  WThat  is  the  value 
of  the  generator  voltage.  In  Fig.  250  <})=cos-1  (.7)  and  EC=CB 
is  the  voltage  at  the  receiving  circuit,  EL=AC  is  the  drop  in  the 


584 


PRACTICAL  MATHEMATICS 


line,   and  Eg  =  AB   is   the   generator   voltage   and   therefore   (1) 
follows : 

(1)  Eg  *»  V(EL  +Ec  cos  4>) 2  +  (Ec  sin  4>) 2  = 


R  i=  <.5  ohm9 


E  =  10  000"V. 
■  P.F.=.7 

|  100  Amps. 


Fig.  257. — A  Transmission  Line. 

Ex.  41.  Determine  the  generator  voltage  for  the  transmission 
line  described  in  Ex.  40  when  the  receiving  circuit  has  the  follow- 
ing values  for  its  power  factor:  (a)  .9;  (6)  .95;  (c)  .85;  (d)  .8; 
(e)  .75;  (/)  .92;  (g)  .93;  (h)  .94;  (j)  .96;  (k)  .97;  (I)  .975;  (m) 
.91;   (n)  .945. 

Ex.  42.  Determine  the  drop  across  two  condensers  of  negli- 
gible resistance,  which  are  connected  in  series  to  110- volt  A.C. 
mains:  (a)  Ci=.5  mf.,  C2  =  .25  mf.;  (6)  d=.5  mf.,  C2=.01  mf.; 
(c)  Ci=.05  mf.,  C2  =  .5  mf.;  (d)  what  is  the  effect  upon  the  drop 
if  the  voltage  is  increased  to  1100  volts;  (e)  what  is  the  effect 
upon  the  respective  drops  if  both  condenser  capacities  are  doubled. 

Ex.  43.  An  electrostatic  voltmeter  has  a  capacity  of  .0006  mf. 
when  it  deflects  with  75  volts  at  its  terminals.  An  auxiliary 
condenser  of  .001  mf.  is  connected  in  series  with  it.  What  voltage 
across  the  combination  will  produce  a  like  deflection? 

Ex.  44.  An  electrodynamometer  has  a  resistance  of  1650  ohms 
and  an  inductance  of  .0205  henry  gives  the  same  deflection  for  a 
60-cycle  E.M.F.  of  unknown  value  as  it  does  for  a  D.C.  voltage 
of  125  volts.    What  is  the  effective  value  of  the  60-cycle  E.M.F.? 

Ex.  45.  An  alternator  delivers  current  to  two  receiving  circuits 
A  and  B  in  parallel.  A  has  a  power  factor  of  .8  and  takes  25  amps. 
B  has  a  power  factor  of  .7  and  takes  20  amps.  What  is  the  power 
factor  of  the  combination?  Construct  the  vector  diagram  by 
laying  off  E  the  E.M.F.  as  a  reference  line,  and  then  the  currents 
in  A  and  B  will  lag  behind  E.  Their  resultant  will  be  the  total 
current.    The  total  phase  angle  is  expressed  in  (1): 


(i) 


Ib  cos  $b+Ia  cos  <Jm 

COS  <|>total  = r 

1  total 


ALTERNATING  CURRENT  PROBLEMS 


585 


Ex.  46.  The  receiving  circuit  is  the  same  as  described  in 
Ex.  45,  excepting:  (a)  A  takes  20  amps.;  (b)  A  takes  30  amps.; 
(c)  B  takes  25  amps.;  (d)  B  takes  15  amps.;  (e)  B  has  a  P.  F.  =.9; 
(/)  B  has  a  P.F.  =.6;  (g)  B  has  a  P.F.  =.95;  (h)  A  has  a  P.F.  =.9; 
(i)  A  has  a  P.F.  =.7;   (j)  A  has  a  P.F.  =.95;   (ib)  A  has  a  P.F.  =1. 

Ex.  47.  An  alternator  delivers  a  60-cycle,  1100- volt,  200-amp. 
current  to  a  receiving  circuit.  What  capacity  would  be  required 
to  compensate  for  the  lagging  current  when  the  power  factor  is: 
(a)  .9;  (b)  .8;  (c)  .95;  (d)  1;  (e)  when  the  P.F.  =.9  and  the  fre- 
quency =  25  ~? 

Ex.  48.  The  power  P  in  an  A.C.  circuit  may  be  measured 
directly  with  a  wattmeter.  It  may  be  measured  also  by  connect- 
ing a  resistance  R  in  series  with  the  circuit  whose  impedance  is 
represented  by  D,  as  shown  in  Fig.  258.  Three  voltmeters  whose 
readings  are  E,  Ei,  and  E2  are  connected  between  the  points  C 
and  A,  B  and  A,  C  and  B,  respectively.  The  working  equation 
is  given  in  (1).  Determine  the  power  in  the  circuit  when  #  =  110, 
tfi  =80,  E2  =40,  and  R  =  10O: 

(1)  P=^(E^-E^-E,J). 


CL'ooooooaoooo'> — /WWWA-ja 
!« E2 >k— Ex — m 


Fig.  258.— Three- Voltmeter 
Method. 


i3 

O 


— O^w/vww 


Fig.  259.  —Three-Ammeter 
Method. 


Ex.  49.  Another  method  of  measuring  the  power  in  an  A.C. 
circuit  is  by  connecting  a  resistance  R  in  shunt  with  the  circuit 
whose  impedance  is  represented  by  D,  as  shown  in  Fig.  259. 
Three  ammeters  whose  readings  are  /,  h,  and  h  are  connected 
in  the  main  and  in  the  branches  D  and  R  respectively.  The 
working  equation  is  given  in  (1).  Determine  the  power  in  the 
the  circuit  when  /  =10.2,  h  =8.1,  I*  =5.5,  and  R  =2012: 


(1) 


P=|(/2_/12_/22)< 


586 


PEACTICAL  MATHEMATICS 


Ex.  50.  A  balanced  two-phase  system  has  a  common  return 
wire.  Determine  the  current  in  the  return  wire  when  each  phase 
has  a  current  of:  (a)  150  amps.;  (6)  200  amps.;  (c)  225  amps. 
What  is  the  current  in  each  phase  when  the  current  in  the  return 
wire  is:   (d)  175  amps.;   (e)  150  amps.;   (/)  200  amps. 

(1)  Ia=Ib  =200  amps. 

(2)  /  =  VIa2+Ib2=IaV2  = 


200  A 


200  A 


>*200A 


Fig.  260. — Balanced  Two-Phase  System  with  Common  Return. 

Ex.  51.     A  balanced  two-phase  system,  Fig.  261,  with  a  com- 
mon return  has  an  E.M.F.  of  110  volts  in  each  phase,     (a)   What 


o 


1-2     Phase  A 
2-3        ••        B 


Fig.  261. — Balanced  Two-Phase  System  with  Common  Return. 

is  the  E.M.F. ,  E,  between  outside  wires.  If  the  two  windings  of 
the  machine  are  connected  in  series  what  E.M.F.  will  the  generator 
develop? 

(1)  EA  =#2*  =  110  volts. 

(2)  E  =  VEa2+Eb*=EaV2  = 


< 

< 

J^ 

i 

ii 

g   < 

o 

/x_ 

I    N 

o 

L 

o 
o 

Fig.  262.— Balanced  Threr-Phase  System. 

Ex.  52.     Fig.  262  represents  a  three-phase  alternator  with  four 
collecting  rings  connected  to  the  four  mains  1,  2,  3,  4,  respectively. 


ALTERNATING  CURRENT  PROBLEMS 


587 


Three  receiving  circuits,  a,  b,  c,  each  take  200  amps,  (a)  What  is 
the  current  in  main  4?  (6)  If  the  connections  for  coil  A  are  reversed 
show  that  the  vector  diagram  changes  from  P'ig.  263  to  Fig.  264. 
What  is  the  value  of  the  current  in  main  4?  (c)  If  the  connections 
for  coils  A  and  B  are  both  reversed  show  that  the  vector  diagram 
changes  from  Fig.  263  to  Fig.  265.  What  is  the  value  of  the  current 
in  main  4?  (d)  If  the  three  windings  of  the  generator  are  connected 
in  series,  what  E.M.F.  will  the  machine  develop? 


Fig.  263. 


Fig.  264. 


Fig.  265. 


Ex.  53.  Fig.  259  represents  three  similar  receiving  circuits 
which  are  A-connected  to  three  phases  between  each  pair  of  which 
there  is  an  E.M.F.  =1100  volts.  In  a  balanced  system  the  total 
power  equals  the  power  in  each  phase  multiplied  by  the  number 
of  phases.     The  total  power  =  150  K.W.,  what  is  the  value  of  the 


266. — A-Connection. 


current  in  each  main  and  the  current  in  each  receiving  circuit 
when  the  power  factor  of  each  receiving  circuit  equals:  (a)  .96; 
(6)  .95;  (c)  .94;  (d)  .93;  («)  .92;  (/)  .91;  [g)  .9;  (h)  what  is  corre- 
sponding effect  upon  the  current  values  when  the  total  power  is 
reduced  to  100  K.W.? 

Ex.  54.  In  Ex.  53  substitute  a  Y-connection  for  the  three 
receiving  circuits  and  determine  the  corresponding  currents  in 
the  mains  and  in  each  receiving  circuit  when  the  power  factor  of 


583  PRACTICAL  MATHEMATICS 

each  of  the  latter  equals:    (a)  .97;    (b)  .98;    (c)  .96;    (d)  .95;    (e) 
.94;   (/)  .93;   (g)  .92;   (h)  .91;   (i)  .9. 

Ex.  55.  Fig.  267  represents  a  three-phase  A-connected  gen- 
erator supplying  three  similar  Y-connected  receiving  circuits. 
Determine  (I)  the  current  in  each  receiving  circuit,  (II)  the 
current  in  each  main,  (III)  the  current  in  each  armature  winding, 
(IV)  the  voltage  in  each  receiving  circuit,  (V)  the  voltage  between 
mains.  Each  armature  winding  develops  an  E.M.F.  =110  volts. 
The  receiving  circuits  have  the  following  respective  resistances 
and  reactances:  (a)  R=5il,  X=2.5S2;  (6)  #=4l2,  X=2.5ft;  (c) 
R=3n,  X  =2.512;  (d)  R=2.5Q,  X=2.5i2;  (e)  #=5i2,  X=3l2; 
(/)  R=5Q,  X=4l2;  {g)  R=5n,  X=5i2. 


Fig.  267. — Three-Phase   A-Connected    Generator  Supplying  Y- Con- 
nected Receiving  Circuit. 

Ex.  56.  A  three-phase  Y-connected  generator  supplies  three 
similar  A-connected  receiving  circuits.  Determine  (I)  the  current 
in  each  receiving  circuit;  (II)  the  current  in  each  main;  (I IT) 
the  voltage  between  mains;  (IV)  the  current  in  each  armature 
winding;  (V)  the  voltage  in  each  receiving  circuit.  Each  armature 
winding  develops  an  E.M.F.  =110  volts.  The  receiving  circuits 
have  the  following  respective  resistances  and  reactances:  (a) 
R=5Sl,  X=2.5<2;  (6)  R=4n,  X=2.5i2;  (c)  fl=3.5i2,  X=2a; 
(d)  #=2.50,  X=25S2;  (e)  R=fo,  Z=3.5S2;  (/)  R=5il,  Z=4l2; 
(g)  R=5Q,  X=5l2. 

Ex.  57.  Make  a  sketch  of  a  three-phase  A-connected  generator 
with  the  mains  supplying  a  A-connected  receiving  circuit.  Repre- 
sent two  wattmeters  Wx  and  W 2  so  connected  that  their  ammeter 
coils  (series  coils)  are  inserted  directly  in  mains  1  and  3  respectively. 
Connect  the  voltmeter  coils  of  Wi  between  mains  1  and  2  and 
the  voltmeter  coils  of  W2  between  mains  2  and  3.  The  total 
power  delivered  to  the  three  circuits  is  25  K.W.  Draw  the  vector 
diagram  for  current  and  voltages  and  determine  the  reading  of 
each  instrument  when  each  receiving  circuit  has  a  power  factor 
equal  to:  (a)  .9;  (b)  .8;  (c)  .85;  (d)  .95;  (e)  1.  Describe  the 
conditions  which  would  result  if  the  receiving  circuits  were  Y- 
connected. 


ALTERNATING  CURRENT  PROBLEMS     589 

Ex.  58.  Two  alternators  A  and  B  are  connected  in  series. 
Their  E.M.Fs.  are  1200  and  1000  volts  respectively.  Plot  the 
power  curves  for  each  generator  and  also  for  the  copper  losses 
in  the  receiving  circuit,  when  the  resistance  and  reactance  of  the 
latter  are:  (a)  #  =  ln,  X  =  1.5ft;  (b)  R  =  ln,  X  =  ltt;  (c)  #  =  1.5n, 
X  =  1.5fl;  (d)  R  =  1Q,  X=2a 

Ex.  59.  An  alternator  is  operated  as  a  synchronous  motor 
when  connected  to  60  ~  1100-volt  mains  and  takes  a  leading 
current.  Its  resistance  =  1.212  and  its  reactance  =  .612.  Determine 
the  following  facts  for  zero  load:  (a)  the  value  of  the  current; 
(6)  the  component  of  the  current  which  is  90°  ahead  of  the  E.M.F. 
of  the  supply;  (c)  the  capacity  of  condensers  which  could  be  sub- 
stituted so  as  to  take  the  same  amount  of  leading  current. 

Ex.  60.  A  synchronous  motor  A  and  an  induction  motor 
B  are  operated  in  parallel  from  110- volt  mains.  Determine  the 
generator  voltage  when  A's  power  factor  =  .92  and  B's  power 
factor  =  .85.  A  takes  a  leading  current  of  50  amperes  and  B 
takes  a  lagging  current  of  85  amperes.       % 

Ex.  61.  A  25  ~  220-volt  generator  delivers  100  amperes 
to  operate  a  number  of  induction  motors  whose  power  factor 
is  .8.  What  capacity  must  be  inserted  to  raise  the  power  factor 
to:  (a)  .85-  (b)  .9;  (c)  .95? 


CHAPTER  XXXIIT 
THE  ALGEBRA  OF  A  TRANSFORMER 


1.  A  transformer  is  an  electrical  device  for  raising  or 
lowering  potential  and  accordingly  is  known  as  a  step-up 
or  step-down  transformer.  It  consists  of  a  core  of  laminated 
soft  iron  or  annealed  sheet  steel  on  which  two  insulated  coils 
are  wound.  The  coil  which  is  connected  to  the  supply 
end  of  the  circuit  is  called  the  primary  whereas  the  coil 
which  is  connected  to  the  receiving  circuit  is  called  the 
secondary.  Transformers  are  classified  as  core-types  when 
the  coils  are  wound  around  the  cores  and  as  shell-types 
when  the  iron  core  is  built  around  the  coils  as  illustrated 
in  cross-section  in  Figs.  269  and  276.  Transformers  are 
represented  diagrammatically  in  Figs.  270  to  275. 

When  an  alternating  current  is  supplied  to  the  primary 
coil  of  a  transformer  it  causes  an  alternating  magnetic  flux 

through  the  core,  which  in  turn 
induces  an  alternating  E.M.F.  in 
the  secondary,  which  produces 
the  alternating  current  in  the 
receiving  circuit. 

2.  The  stationary  coils  of  a 
transformer  with  their  correspond- 
ing periodically  varying  fields  may 
be  likened  to  a  coil  rotating 
through  a  uniform  field  as  shown 
in  Fig.  268.  Therefore  the  formu- 
las for  the  induced  E.M.F.  of 
a  generator  apply  to  the  induced  E.M.Fs.  of  a  transformer. 

590 


/                                   \ 

'*t;<>. 

\    "^v 

illliltfi 

Fig.  268. 


THE  ALGEBRA  OF  A  TRANSFORMER.     591 

The  rapid  reversals  of  magnetization  in  the  iron  core 
induces  an  E.M.F.  e  in  each  turn  of  both  primary  and 
secondary  coils  and  if  their  respective  number  of  turns  be 
represented  by  JVi  and  N2  then  the  corresponding  total 
E.M.F.  is  expressed  by  (1)  and  (2). 

(1)  Ei=Nie, 

(2)  E2  =  N2e, 

(3)  •      E±-*± 

The  interpretation  of  (3)  states  that  the  ratio  of  the 
primary  and  secondary  E.M.Fs.  equals  the  ratio  of  their 
respective  number  of  turns. 

The  magnetizing  action  of  the  secondary  current  I2  is 
neutralized  by  an  equal  and  opposite  magnetizing  action 
of  the  primary  current  7i  as  expressed  in  (4)  from  which 
(5)  follows. 

(4)  NiIi=Nzl2, 

(5)  *  ■      ^  =  ^ 

^  ■■  5"n- 

(6)  results  from  equating  (3)  and  (5)  and  interpreted 
states  that  the  currents  in  the  primary  and  secondary 
coils  vary  inversely  as  their  respective  E.M.Fs.  In  an 
ideal  transformer  the  active  power  I22R2  of  the  secondary 
should  have  an  equal  equivalent  primary  power  I\2R  in 
terms  of  the  primary  current  as  expressed  in  (7). 

(7)  h2R=I*>R2, 

(8)  /.     R=(fy2R2, 


m  ■■■  «=(£) 


2 

R2. 


592  PRACTICAL  MATHEMATICS 

(9)  results  from  the  substitution  of  (5)  in  (8).  In  (9) 
R  is  called  the  equivalent  resistance,  i.e.,  the  resistance 
which  would  have  to  be  placed  in  the  primary  to  give  a 
corresponding  loss  due  to  R2  in  the  secondary. 

The  wattless  power,  h2X2,  of  the  secondary  should 
have  an  equal  primary  power  equivalent  Ii2R  in  terms  of 
the  primary  current  as  expressed  in  (10). 


(10)  h2X  =  I22X2, 


A', 


wJ 


(12)  .-.  x-$y*. 

(12)  results  from  the  substitution  of  (5)  in  (11).  In 
(12)  X  is  called  the  equivalent  reactance,  i.e.,  the  reactance 
which  would  have  to  be  placed  in  the  primary  to  require 
a  corresponding  energy  due  to  X2  in  the  secondary. 

3.  Transformer  Losses  and  Efficiency.  The  total  losses 
Wt  in  a  transformer  are  composed  of  copper  losses  Wc 
due  to  the  resistance  in  the  primary  and  secondary,  hysteresis 
losses  Wn  and  eddy  current  losses  We. 


(13) 

Wc=h2Ri+h222, 

(14) 

Wt-KiVfB1*, 

(15) 

We=K2Vf2B2, 

(16) 

WT=WC+Wh+We. 

(14)  and  (15)  are  identical  with  (37)  and  (38)  respectively 
in  Chap.  IX.  The  efficiency  y]  of  a  transformer  is  expressed 
in  (17)  in  which  W  is  the  input  of  the  transformer. 

output  =  W-Wt 
1  ~  input  W      ' 

Ex.  1.  (a)  What  elements  affect  both  eddy  and  hysteretic 
losses?     (b)  Of  these  elements  which  produce  like  effects  upon 


THE  ALGEBRA  OF  A  TRANSFORMER 


593 


both  losess?  (c)  Of  the  remaining  elements  which  have  the  greater 
effect  upon  these  losses?  (d)  Are  there  any  elements  which 
affect  one  loss  and  not  the  other?  (e)  How  could  (14)  and  (15) 
be  changed  to  express  the  losses  in  terms  of  weight? 

4.  The  Design  of  the  Core.  The  core  losses  in  a  trans- 
former vary  directly  as  the  weight  of  the  core  and  therefore 
the  latter  should  be  minimized. 

Ex.  2.  (a)  What  determines  the  cross-sectional  area  and  what 
determines  the  length  of  the  core? 

(b)  Increase  of  length  has  what  effect  upon  the  copper  losses? 

(c)  The  shape  of  the  cross-section  of  the  core  has  a  consider- 
able influence  upon  the  length  of  the  turns,  therefore  what  is  its 
corresponding  effect  upon  the  resistance  of  the  coils? 

(d)  Compare  the  length  of  turns  required  on  re  tangular  square 
and  circular  cross-sections  of  equal  area. 

(e)  Under  what  circumstances  will  a  circular  cross-section  be 
preferable? 

(/)  If  the  linear  dimensions  are  the  same  what  is  the  ratio 
of  volumes  of  circular  and  square  cores? 

(g)  Which  cross-section  therefore  gives  the  greater  E.M.F. 
induced  per  foot  of  wire?     How  does  the  difference  increase? 

Ex.  3.  Make  a  comparative  study  of  the  different  types  of 
transformers  illustrated  in  Fig.  269.    Assume  a  like  output,  i.e., 


fa] 


(b) 


i 


i: 

<K.» 

<v: 

pan 

Fig.  269. — Types  of  Transformers. 

product  of  current  and  E.M.F.,  for  each  transformer.  An  equal 
current,  current  density,  and  gauge  of  wires  will  be  used  for  all 
windings.  The  number  of  turns  will  be  directly  proportional  to 
the  winding  space  and  for  the  same  induction  the  cross-section 
will  be  inversely  proportional  to  the  number  of  turns.  The 
weight  of  iron  and  the  length  of  wire  may  be  taken  as  a  basis 
to  judge  the  design.    The  dimensions  of  the  figures  are  given  in 


594 


PRACTICAL  MATHEMATICS 


centimeters.  Type  (a)  has  a  core  whose  cross-section  =400  sq.cm. 
and  winding  space  =60  sq.cm.  The  weight  of  the  iron  =200  kg. 
The  mean  length  of  one  turn  =2x40 +2x10 +2x3  =  118.9  cm. 
100  turns  on  the  primary  require  118.9  meters  of  wire. 

Ex.  4.  Calculate  the  eddy  current  and  hysteretic  losses  in 
the  iron  of  a  60-cycle  transformer  for  which  Bm  =  .15  mega- 
maxwells.  The  mean  length  of  the  magnetic  circuit  is  30  ins. 
and  the  cross-section  8  sq.ins. 

Ex.  5.  A  transformer  has  a  primary  winding  of  100  turns  of 
No.  4  B.  &.  S.  copper  wire  and  the  secondary  winding  of  2000  turns 
of  No.  16  B.  &  S.  copper  wire.  The  mean  lengths  of  these 
respective  windings  are  30  ins.  and  18  ins.  Determine  the  total 
equivalent  primary  resistances. 

Ex.  6.  A  110- volt  D.C.  main  supplies  a  current  to  a  coil  of 
200  turns  wound  on  an  iron  core,  as  shown  in  Fig.  270.     The 


r^=i— <-?200      3! 


s|,|, 


H'H'H'I'I'I'I'I'M 

110  Volts  D.C. 


:        ;->n2= 

<—>  <     >   100 


Fig.  270. 

current  is  reversed  120  times  a  second.  The  E.M.F.  in  the  primary 
is  shown  as  curve  Ek  in  Fig.  271.  The  core  flux  is  represented  by 
curve  <J>  in  Fig.  271;  it  has  a  constant  slope  and  reaches  its  max- 
imum and  minimum  values  at  the  instant  of  current  reversals, 
i.e.,  when  the  primary  E.M.F.  is  zero.  The  secondary  E.M.F. 
is  represented  by  E,,  and  is  180°  out  of  phase  with  En  i.e.,  when 
E,  is  positive  Elt  is  negative,  whereas  when  Et  is  negative  E2  is 
positive. 

A  secondary  coil  of  100  turns  is  wound  on  the  above  core  and 
supplies  a  current  to  a  non-inductive  receiving  circuit  =50a 
Both  primary  and  secondary  coils  have  negligible  resistance. 
Determine  the  secondary  E.M.F.,  secondary  current,  total  primary 
current,  core  flux  when  its  reluctance  =  .0001. 


(1) 
(2) 


Si    N* 

E2    Ni 

E2  = 


THE  ALGEBRA  OF  A  TRANSFORMER 
\ 


595 


596 

(3) 

(4) 


PRACTICAL  MATHEMATICS 
dA>     Ei     110X108 


dt     Ni 
4>=55X106* 


200 

55X106 
120 


=  55X106. 


Ex.  7.  A  transformer  has  its  primary  and  secondary  coils 
wound  with  1100  and  110  turns  of  wire  respectively,  as  shown 
in  Fig.  272.     The  supply  current  is  from  1100- volt  mains.   What 


r 


P.F.=  9 


I„=  150  Amps.. 


Fig.  272. 


High  Tension 


R  =  100fi 


is  the  equivalent  resistance  and  equivalent  reactance  when  the 
secondary  delivers  150  amps,  to  a  receiving  circuit  having  a  power 
factor  equal  to:  (a)  .97;  (6)  .96;  (c)  .95;  (d)  .94;  (e)  .93;  (/) 
.92;  (g)  .91;  (h)  .90;  (i)  .89;  (j)  .88;  (k)  .87;  (I)  .86;  (m)  .85; 
(n)  C2  =200  amps,  and  P.F.  =.95. 

Ex.  8.  A  10:1  transformer  has  its  primary  coil  connected  to  a 
1100- volt  main,  as  shown  in  Fig.  273.  The  primary  coil  is  sup- 
plied through  a  variable  non-induc- 
tive resistance,  Ri  =  100tt.  Deter- 
mine the  resistance  and  reactance  of 
a  simple  circuit  which  would  replace 
the  transformer  and  secondary. 
Determine  the  primary  and  second- 
ary currents,  the  primary  and 
secondary  terminal  voltages.  The 
secondary  delivers  current  to  a 
circuit  having  the  following  resist- 
ances and  reactances :  (a)  R2  =  512, 
Xi-2C;  (6)  R2=3a,  X2  =  lQ; 
(c)  #2=3tt,  X2=2tt;  (d)  R2  =  ln, 
X2  =  .5fl;  (e)  R2=0il,  X2=3l2;  (/)  fl2=3fl,  X««30j  (<?)  #2  =2.5*2, 
X2=3i2;  (h)  R2=3n,  X2  =2.512. 

Ex.  9.     An  iron  core  consisting  of  a  bundle  of  wires  of  15  sq.cm. 
cross-sectional  area  is  magnetized  from  an  A.C.  60-cycle  110- volt 


1100 

v. 


ossmi 


[CoTOBTT 


R  =2 


Fig.  273. 


THE  ALGEBRA  OF  A  TRANSFORMER 


597 


main,  (a)  How  many  turns  are  required  to  give  a  maximum 
flux  density  =4000  lines  per  square  centimeter?  How  is  the  result 
altered  when  (6)  the  frequency  =25 ~;  (c)  when  the  E.M.F.  of 
mains  =550  volts;  (d)  when  the  flux  density  =3500  lines  per 
square  centimeter.     See  Fig.  274. 

Ex.  10.  A  step-up  transformer  raises  the  voltage  of  an  alter- 
nator from  110  volts  to  15,000  volts.  There  are  50  turns  in  the 
primary  coil  and  500  cm.  of  cross-section  are  allowed  for  each 
ampere  of  current.  Determine  the  number  of  turns  in  the  second- 
ary and  the  cross-sections  of  primary  and  secondary  windings 
when  the  alternator  supplies:  (a)  500  amps.;  (6)  750  amps.;  (c) 
250  amps.;  (d)  200  amps.;  (e)  100  amps.;  (/ )150  amps.,  Fig.  27QA. 


Iron  Wires 


S 


Fig.  275. 


Ex.  11.  A  transformer,  Fig.  275,  has  1000  turns  in  the  primary 
and  100  turns  in  the  secondary.  The  primary  wire  is  40  mils  in 
diameter  and  500  cir.  mils  are  allowed  per  ampere.  The  sectional 
area  is  10  sq.cm.  Determine  the  E.M.F.,  current,  and  power 
rating  of  the  transformer  when  the  frequency  and  flux  density 
are:  (a)  /  =  60~,  5=4200  lines  per  square  centimeter;  (b)  /  = 
25~,  5=4100  lines;  (c)  /  =  60~,  5=4100;  (d)  /=25~,  5  = 
4000;     (e)  /  =  60~,  5=4000;   (/)  =25~,  5=4200. 

Ex.  12.  Determine  the  efficiency  of  a  10  K.V.A.  transformer 
at  unity  power  factor  whose  iron  loss  constantly  equals  160  watts 
and  whose  load  L  and  copper  losses  Pc  are  as  follows:  (a)  L  =2500. 
Pe  =  10;  (6)L=5000,PC=40;  (c)  L=7500,  Pc=90;  (d)  L  =  10,000, 
Pc  =  160;  (e)  L  =  12,500,  Pc  =  250.     Construct  the  efficiency  curve. 

Ex.  13.  A  shell  type  transformer  is  shown  in  cross-section 
and  in  longitudinal  section  in  Fig.  276.  The  mean  length  of 
both  primary  and  secondary  coils  is  29.5  ins.  The  primary  con- 
sists of  480  turns  of  No.  7  B.  &  S.  copper  wire.  The  secondary 
consists  of  24  turns  of  two  strands  of  No.  7  wires  in  parallel.  The 
section  of  the  magnetic  circuit  is  10x1.75  ins.  .85  of  the  section 
is  iron.     The  volume  of  the  iron  is  24  sq.ins.  X10  X.85  ins.     Thick- 


598 


PRACTICAL  MATHEMATICS 


ness  of  laminations  15  mils.  Allow  500  cir.  mils  per  ampere  and 
a  flux  density  of  4500  lines  per  square  centimeter.  Determine 
(a)  the  E.M.F.,  current,  and  power  ratings  of  the  transformer  at 
60  cycles;  (&)'  the  copper  and  iron  losses  at  full  load;  (c)  the 
efficiency  of  the  transformer  at  full  load;  (d)  the  all-day  efficiency 
of  the  transformer  when  operated  for  4.5  hours  each  day  at  full 
load  and  at  zero  load  for  the  remainder  of  the  day. 


Fig.  276.— Cross-sections  of  a  Shell  Type  Transformer. 

Ex.  14.  A  shunt  motor  has  an  output  of  5  K.W.  At  zero 
load  the  loss  is  850  watts  and  the  losses  in  the  motor  at  full  load 
are  1050  watts.  Determine  the  all-day  efficiency  when  the  motor 
is  operated  at  full  load  intermittently  three  minutes  out  of  every 
ten  minutes  during  ten  hours  of  the  day. 

Ex.  15.  A  10-K.W.  1100  :  110-volt  transformer  is  connected 
as  an  autotransformer  to  step  up  the  voltage  from  1100  volts 
to  1175  volts,  t.  Determine  (a)  the  power  delivered  at  1175  volts 
to  non-reactive  circuit  without  exceeding  the  rating  of  the 
transformer;  (b)  the  current  in  each  coil  of  the  transformer; 
(c)  the  power  delivered  to  and  by  the  coils.  II.  The  transformer 
steps  down  so  as  to  deliver  1000  volts  to  the  non-reactive  circuit. 
III.  The  transformer  steps  up  to  1200  volts.  IV.  The  transformer 
steps  down  to  1025  volts. 

Ex.  16.  Three  1100  :  110-volt  transformers  have  their  1100- 
volt  coils  delta  connected  to  a  three-phase  1100-volt  mains.  The 
secondaries  are  Y-connected  to  service  mains,  as  shown  in  Fig.  277. 


500  Amps. 


Fig.  2764. 


Fig.  277. 


(a)  Determine  the  E.M.Fs.  between  each  set  of  mains.  (6) 
Determine  the  E.M.Fs.  between  each  set  of  mains  when  the  delta 
and  Y-connections  are  interchanged,  as  shown  in  Fig.  278. 


THE  ALGEBRA  OF  A  TRANSFORMER  599 

Ex.  17.     Two  100-K.W.  transformers,  Fig.  279,  are  arranged 
for  the  step-down  transformation  of  a  three-phase  supply.     Deter- 


Fig.  278. 

mine  the  power  which  can  be  delivered  to  balance  the  receiving 
circuits  without  exceeding  the  voltage  and  current  ratings  of  the 
two  transformers  when  the  power  factor  of  the  receiving  circuits 
is:   (a)  P.F.=.9;   (b)  .95;   (c)  .90;   (d)  .85. 


P.  F.=. 


Fig.  279. 

Ex.  18.  Two  transformers  are  connected  to  two-phase  mains. 
Each  phase  has  an  E.M.F.  =  1000  volts.  The  primary  coils  have 
each  480  turns.  The  secondaries  are  joined  in  series  and  give 
an  E.M.F.  =30°  ahead  of  one  of  the  two  E.M.Fs.  Determine 
the  proper  winding  for  each  secondary  coil  when  its  E.M.F.  equals: 
(a)  550  volts;  (6)  600  volts;  (c)  500  volts. 

Ex.  19.  A  Scott  transformer  is  to  transform  from  1100- volt  60- 
cycle  two-phase  to  a  110-volt  three-phase.  The  cross-section  of 
the  core  is  70  sq.cm.  The  maximum  flux  density  =4000  lines  per 
square  centimeter.  I.  Determine  (a)  the  number  of  turns  of  wire 
in  each  primary  coil;  (6)  the  number  of  turns  of  wire  in  each 
secondary  coil.     II.  Determine  (a)  and  (6)  when/=25~. 

Ex.  20.  A  three-ring  converter  is  to  supply  a  D.C.  to  a  car 
line  at  600  volts.  Three  similar  transformers  are  provided  to 
accomplish  the  step-down  transformation  from  10,000  volts.  The 
primaries  are  delta  connected  to  the  high  voltage  mains  and  the 
secondaries  are  Y-connected  to  the  three  rings  of  the  converter. 
What  is  the  ratio  of  transformation  for  each  transformer  when 
the  voltage  of  the  car  line  equals:  (a)  600  volts;  (6)  575  volts; 
(c)  550  volts? 


600 


PRACTICAL  MATHEMATICS 


Ex.  21.  The  thirty-six  windings  of  an  ordinary  ring-wound 
D.C.  armature  are  disconnected  from  the  commutator  and  num- 
bered consecutively  from  1  ...  36.  Specify  and  illustrate  with 
a  diagram  the  manner  in  which  the  thirty-six  coils  should  be 
connected  to  two-phase  60-cycle  mains  in  order  to  produce  in 
the  ring  a  rotating  state  of  magnetism  of  (a)  4  poles,  (6)  6  poles, 
(c)  12  poles,  (d)  18  poles,  (e)  state  the  speed  of  the  magnetism  in 
each  in  revolutions  per  second. 

Ex.  22.  An  ideal  three-phase  induction  motor  takes  5  amps, 
into  each  phase  of  its  delta  connected  primary  member  at  110  volts 
60  cycles.  The  rotor  runs  at  two-thirds  synchronous  speed.  Deter- 
mine (a)  the  ratio  of  the  stator  to  the  rotor  turns;  (b)  the  rotor 
terminal  voltage;  (c)  total  intake  of  power;  (d)  electrical  output 
of  power;  (e)  the  mechanical  output  of  power.  The  rotor  which 
has  a  three-phase  winding  delta-connected  to  collector  rings  supplies 
15  amps,  to  each  of  three  similar  circuits,  each  having  a  power 
factor  equal  to:   (a)  .9;   (6)  .85;   (c)  .8;   (d)  .75. 


2.0- 


I=V 

E 

u 

I 
< 

=rM27r/L--^ 

I 

DATA 

E  =110  VOLTS 

R  =2  OHMS 

L  =  .352  HENRYS 

C  —  20  MICROFARADS 
/  — 0 120  CYCLES 

"J^J/ 

zi.0- 

1- 
z 
M 
tc 

BE 

O    .5- 

T^ 

> c 

! { 

I        I 

10   20        30   40    50   60   70    80   90   100   110   120 

frequency 

Fig.  2794. 

Ex.  23.  Make  the  necessary  calculations  and  construct  the 
curve  shown  in  Fig.  279A.  Plot  between  current  and  frequency. 
Use  the  data  given  for  the  figure. 


CHAPTER  XXXIV 
THE  CIRCLE   DIAGRAM 

1.  Variable  Elements  in  an  A.C.  Circuit.  In  the  pre- 
ceding chapters  A.C.  problems  were  solved  by  numeric  cal- 
culation and  also  by  vector  diagrams  when  a  sufficient  and 
necessary  amount  of  data  was  specified.  When  any  data 
includes  definite  values  of  resistance,  reactance,  frequency 
and  impressed  E.M.F.  there  is  a  corresponding  definite 
vector  diagram  and  its  corresponding  impedance  or  admit- 
tance diagram.  Every  new  set  of  data  implies  its  corre- 
sponding diagrams.  In  this  chapter  a  general  diagram 
called  a  circle  diagram  or  a  locus  diagram  will  be  described 
by  means  of  which  it  will  be  possible  to  study  the  variation 
of  currents,  E.M.Fs.,  power  factors,  phase  angles,  power, 
etc.,  in  a  circuit  which  is  subject  to  variable  resistance, 
reactance,  and  frequency.  Fig.  280  shows  the  usual  vector 
diagram  for  a  circle  supplied  with  constant  E.M.F.  E 
and  in  which  the  resistance  and  reactance  have  the  constant 
values  R  and  X  respectively.  AB  represent  the  constant 
E.M.F.  E,  and  AI  represents  the  current  /  -  lagging  0° 
behind   the   E.M.F.     AD  =  IR  =  ER  =  the   drop   across   the 

resistance  R  whereas   DB  =  IX  =  Ex  =  ihe    drop  across  the 

if 
reactance  X  and  0  =tan_1  — .     From  the  property  of  a  circle 

a  right  angle  is  inscribed  in  a  circle.     A  circle  may  be  drawn 
upon  A  B  as  a  diameter  so  as  to  pass  through  D. 

E 


(1)  /  = 


Vr2+x2' 


(2)  /.     E  =  IVR2+X2. 

601 


602 


PRACTICAL  MATHEMATICS 


(1)  and  (2)  express  the  relation  of  four  quantities.  If 
we  consider  E  constant  then  it  is  possible  to  change  the 
current  by  .changing  either  the  resistance  or  the  reactance 
as  shown  in  (3)  and  (4)  respectively,  or  by  changing  both 
the  resistance  and  the  reactance  as  shown  in  (5).     Hereafter 

IR 


AAAAA/VWWVWVVQ 


Fig.  280. — The  Diagram  for  a  Circuit  having  Constant  Elements. 


constant  values  are  represented  by  capital  letters  and  variable 
or  instantaneous  values  are  represented  by  1.  c.  letters. 


(3) 
(4) 
(5) 


E  =  i\/r2+X2, 
E  =  iVR2+x2, 


E  =  iVr2+x2, 


In  the  above  cases  E  is  always  the  hypotenuse  of  a  right 
triangle  and  the  mutually  orthogonal  components  ER  and 
Ex  are  the  legs.  The  right  angle  of  each  right  triangle 
will  lie  on  the  circumference  of  a  circle  described  on  the 
hypotenuse  as  a  diameter.  The  circle  is  the  locus  of  the 
point  D  for  circuits  in  which  E  is  constant  and  i,  r,  and  x 
are  variable.  The  lower  semicircle  is  used  when  the  phase 
angle  6  is  positive,  i.e.,  when  the  inductive  reactance  is  in 


THE  CIRCLE  DIAGRAM 


603 


excess.     The  upper  semicircle  is  used  when  0  is  negative, 
i.e.,  when  the  capacity  reactance  is  in  excess. 

2.  Fundamental  Principles  in  an  A.C.  Circuit.  (I)  The 
voltage  E  which  is  impressed  upon  a  circuit  may  be  resolved 
into  two  mutually  orthogonal  components,  the  one  ER 
being  in  phase  with  the  current  and  the  other  Ex  being 
normal,  i.e.,  perpendicular  to  the  current. 

(II)  The  current  in  a  circuit  may  be  resolved  into 
two  mutually  orthogonal  components,  the  one  I0  being  in 
phase  with  the  applied  voltage  and  the  other  Ib  being 
normal  to  the  voltage. 

3.  A  circuit  having  constant  resistance  R  and  variable 
reactance  x  with  constant  impressed  voltage  E  is  represented 


/wvvwwww 


Fig.  281. — Circle  Diagram  for  a  Circuit  with  Variable  Reactance. 


in   Fig.    281.      In    triangle   BDC,    <n  =  tan 


The  line 


DC  =  ex  will  vary  with  both  i  and  x  whereas  the  line  BD  =  eR 
is  proportional  to  current  i  only  since  R  is  constant.  ex  is 
represented  in  a  new  position  SC  and  eR  in  the  corresponding 

new  position  BS  and  $2  =  tan-1  —  =  tan-1  — .  By  choosing 

eR  R 

a  suitable  scale  the  line  eR  may  represent  the  successive 

instantaneous  values  of  the  current  i  as  a;  changes.     The 

active  power  W  =  i?R  in  the  circuit  is  proportional  to  BD2. 


WozBD2, 


604 


PRACTICAL  MATHEMATICS 


but 

(6) 
(7) 


BD2  =  BCXBP  =  EXBP, 
.     BD2ozKP     and     WozRP. 


In  (6)  E  =  BC  is  a  constant.  The  interpretation  of  (7) 
states  that  the  active  power  in  the  circuit  is  proportional 
to  and  represented  by  BP  the  projection  of  ER  on  E.  By 
choosing  a  suitable  scale  for  BP  the  power  in  the  circuit 
may  be  read  directly  from  the  circle  diagram. 


(8) 

(9) 


.     BD     BD 
C0S^  =  BC=~T 

cos  fyocBD. 


The  interpretation  of  (9)  states  that  the  power  factor 
of  the  circuit  is  proportional  to  BD.  Since  the  maximum 
value  of  the  chord  BD  equals  the  diameter  E,  then  the 
scale  for  reading  power  factors  may  be  determined  from 
the  fact  that  a  unity  power  factor  is  represented  by  a  line 
equal  to  BC. 

Ex.  1.  Explain  the  method  for  determining  the  scale  when 
BD  in  Fig.  282  is  used  for  reading  the  values  of  the  current  in  the 


Fig.  282. 

circuit.     Explain  why  BC  cannot  represent  the  current. 

Ex.  2.  Explain  the  method  for  determining  the  scale  when 
BP  is  used  for  reading  the  values  of  the  active  power  in  the  circuit. 
What  line  in  the  figure  represents  wattless  power  and  what  is  its 
scale. 

Ex.  3.  Construct  a  circle  diagram,  Fig.  282,  for  a  circuit 
having  constant  E  and  X  but  variable  i  and  r.     In  such  a  diagram 


THE  CIRCLE  DIAGRAM 


605 


state   which  lines  represent   current,   power  factor,   and   active 
power  and  give  sufficient  proof. 

Ex.  4.  Construct  the  circle  diagram  for  Ex.  3,  when  the 
capacity  reactance  is  in  excess. 

Observation.  In  the  circle  diagram  for  constant  voltage 
circuits  the  active  power  line  is  the  horizontal  projection 
of  the  current  line  when  R  is  constant  but  it  is  the  vertical 
projection  of  the  current  line  when  X  is  constant. 

4.  The  Horizontal  Auxiliary  Semicircle.  It  was  observed 
in  a  preceding  chapter  that  when  a  series  circuit  contains 
a  number  of  impedances  the  latter  could  be  separated 
and   rearranged   by   combining  the  individual   resistances 


Fig.  283. — A  Series  Circuit  with  Variable  Reactance. 

into  one  group  and  the  individual  reactances  into  a  second 
group.  The  algebraic  sum  of  the  drops  across  the  individual 
resistances  equals  the  drop  across  the  group  of  resistances, 
and  further  the  algebraic  sum  of  the  drops  across  the 
individual  reactances  equals  the  drop  across  the  group  of 
reactances.  These  principles  are  applied  to  the  circle 
diagram  for  a  series  circuit.  The  circuit  diagram  in  Fig. 
283  shows  two  constant  resistances  R\  and  R2  in  series 
with  a  variable  reactance  xi,  which  are  supplied  with  current 
from  a  source  of  constant  impressed  E.M.F.  E.  The 
equation  for  the  current  in  the  circuit  may  be  written  as 
(10)  irrespective  of  whether  R\  and   x\  are  considered  sep- 


606  PRACTICAL  MATHEMATICS 

arately  or  as  constituting  the  two  elements  of  an  impedance 
*i  =  \/fli2+zi2. 

(10) 


(11)  tanG 


V(Ri+R2)2+xi2' 
ix\  x\ 


iRi+iR2     R1+R2 


The  circle  diagram  in  Fig.  283  is  constructed  so  that 
the  diameter  AB  =  E  and  then 

_XBD  _1      xi 


0  =  tan     — —  =  tan 


AD  R1+R2 

BD  represents  the  instantaneous  drop  across  the  variable 
reactance  x\  and  AD  represents  the  instantaneous  drop 
across  the  total  resistance  R1+R2.  Since  R\  and  R2  are 
constant  their  respective  drops  are  proportional  to  their 
resistances. 

(    .  AF=iR2^R2 

r.  }  FD    iR!    Ri 

It  is  necessary  to  be  able  to  distinguish  the  segments 
of  AD  which  represent  the  respective  drops  across  Ri 
and  R2  for  each  new  position  of  AD.  Draw  FT  1  AF, 
intersecting  AB  at  T.  Construct  the  horizontal  auxiliary 
semicircle  AF\FT,  passing  through  A,  F,  and  T.  The 
point  F  travels  around  the  arc  TFF\A  as  arm  AD  rotates 
about  A,  i.e.,  as  the  point  D  travels  around  the  arc  BDD\A. 
Therefore  the  points  F  and  D  will  always  represent  the 
right  angled  vertexes  of  two  similar  right  triangles. 
Accordingly  the  auxiliary  semicircle  TFFiA  will  continually 
divide  the  rotating  arm  AD  into  proportional  segments 
whose  ratio  is  R2  :  R\,  as  expressed  in  (13). 

t\Vi  AF  =  AT=AFi  =iR2_R2 

U  ;  FD    TB    F1D1    iRi    R{ 


THE  CIRCLE  DIAGRAM 


607 


0  is  the  angle  of  lag  for  the  entire  circuit,  whereas  0i 
is  the  angle  of  lag  for  the  impedance  z\  =  \/Ri2+xi2. 

Fig.  284  is  an  alternative  construction  in  which  the 
#2  and  Ri  drops  are  interchanged  in  position.  Its  dis- 
advantage lies  in  the  fact  that  no  line,  in  the  figure  represents 
the  drop  across  the  impedance  z\. 

Ex.  6.  In  Fig.  284  what  lines  represent  current,  power  factor, 
power  expended  in  Rlt  power  expended  in  R2,  the  active  power 
expended  in  the  circuit,  and  the  wattless  power  in  the  circuit. 

Ex.  6.  Construct  the  circle  diagram  for  the  circuit,  shown  in 
Fig.  284,  wherein  xx  represents  an  excess  of  capacity  reactance, 


Fig.  284. — An  Alternative  Construction. 

R\  =  5ft  ,R2  =  1012,  Xi  ranges  from  2  to  15ft.  What  are  the  maximum 
and  minimum  values  of  the  phase  angles,  power  factors,  and  power 
consumed  in  Rh  R2,  and  Ri+R2.  Determine  these  values  graph- 
ically from  the  circle  diagram. 

5.  The  Construction  of  the  Phase  Angle.     The  phase 


angle    0  =  tan 


Xi 


may  be    readily    constructed    by 


R1+R2 

reducing  the  ratio  to  its  simplest  terms  and  applying  the 
principle  shown  on  page  429  and  therefore  without  any 
knowledge  of  the  degree  measure  of  the  angle.  For  the 
maximum  value  of  the  phase  angle  in  Ex.  6, 

xi  15 

tan  0i  = 


R1+R2     10+5 


1. 


Construct  BJ LAB,  and  =AB.     Join  A  with  «/,  then 
tan  5^LJ  =  l=tan  0i. 


608  PRACTICAL  MATHEMATICS 

For  the  minimum  value  of  the  phase  angle, 

2 
tanGi  =  — =  .133. 
15 

Construct  BKlAB  so  that  BK  =  .133  AB.  Join  A  with  K, 
then  tan  BAK  =  ASS=tsai  0i.  It  is  advantageous  to  con- 
struct the  circle  diagram  on  cross-section  paper  or  polar 
paper  so  that  its  diameter  is  10  units  in  length.  Its  scale 
is  adjusted  accordingly. 

6.  The  Vertical  Auxiliary  Circle.  The  circuit  diagram 
in  Fig.  285  shows  a  constant  resistance  R2,  a  constant 
reactance  Xi  and  a  variable  resistance  n,  which  are  supplied 
with  current  from  a  source  of  constant  E.M.F.  E.  The 
equation  for  the  current  in  the  circuit  may  be  written 
irrespective  of  whether  Xi  and  ri  are  considered  separately 
or  as  constituting  the  two  elements  of  the  impedance 
zi  -  Vri2+Xi2. 

The  circle  diagram  in  Fig.  285  is  constructed  so  that 
the  diameter  AB  =  E  and  then 

,  BD  _!     Xi 

9  =  tan      — — =tan 


AD  n+R2 

BD  represents  the  instantaneous  drop  across  the  constant 
reactance  Xi  and  AD  represents  the  drop  across  the  total 
resistance  n  +-R2.  The  line  AD  is  divided  into  two  segments 
AF=iR2  and  FD  —  ir\.  AF  varies  with  i  alone  whereas 
FD  varies  with  both  i  and  n .  Therefore  for  each  new  position 
of  AD  its  segments  will  bear  a  different  ratio,  i.e.,  AF 
must  vary  with  i  alone.  Construct  AT  LAB  and  FTlAF. 
Through  the  points  A,  F,  and  T  construct  the  vertical 
auxiliary  circle  AFF\T.  The  point  F  travels  around  the 
arc  AFF\T  as  the  arm  AD  rotates  about  A,  i.e.,  as  the 
point  D  travels  around  the  arc  BDD\A.  Therefore  the 
points  F  and  D  will  always  represent  the  right  angled  vertexes 
of  two  similar  right  triangles.     Accordingly  the  segment 


THE  CIRCLE  DIAGRAM 


609 


AF  of  the  rotating  arm  AD  which  is  intercepted  by  the 
arc  AFFiT  is  proportional  to  the  current  i  only  as  expressed 
in  (14).     The  two  diameters  AB  and  AT  are  constant. 


(14) 

(15) 
(16) 


AF 
AT 


DB 
AB' 


AFocBD    but    BDozi, 
.'.     AFoci. 


Therefore  AF  varies  with  i  alone  and  represents  the 
instantaneous  drop  across  the  constant  resistance  R2 


Fig.  285. — A  Series  Circuit  with  Variable  Resistance. 

Ex.  7.  Which  two  lines  of  Fig.  285  may  be  used  as  the  cur- 
rent lines  of  the  diagram.  Shcnv  which  lines  represent  the  several 
power  factors  of  the  circuit,  and  its  elements,  the  several  phase 
angles,  and  the  several  active  powers  in  the  circuit  and  its  elements. 

7.  The  Eccentric  Circle.  An  alternative  construction 
for  the  preceding  problem  is  shown  in  Fig.  286  in  which 
AB  =  E  is  the  usual  diameter  and  FB  is  the  drop  across 
the  constant  reactance  X\  and  AF  is  the  drop  across  the 
total  resistance  ri+ife. 

AF  is  divided  into  two  segments  AD=ir\  and  DF=iR2. 
Therefore  DF  varies  with  current  only.  Construct  the 
circular  arc  ADB  which  is  known  as  the  arc  of  the  eccentric 
circle  whose  center  H  may  be  determined  by  the  intersection 
of  the  perpendicular  bisectors  of  AB  and  AD.      The  point 


610 


PRACTICAL  MATHEMATICS 


F  travels  around  the  arc  BDA  as  the  arm  AF  rotates  about 
A,  i.e.,  as  the  point  D  travels  around  the  arc  BFF\A.  There- 
fore the  points  F  and  D  are  the  vertexes  of  two  inscribed 
angles  which  intercept  constant  arcs  in  their  respective  circles. 
Therefore  the  triangles  DBF  and  D\BF\  are  similar. 

D1F1  =  FiB  =  i1 

FB     %2       • 


(17) 


DF 


Therefore  DF  the  segment  of  the  chord  AF  which  is 
intercepted  between  arcs  BDD\A  and  BFFiA  will  vary 
with  current  only. 


Fig.  286. — The  Eccentric  Auxiliary  Circle. 

Ex.  8.  Which  lines  in  Fig.  286  represent  current,  power  factor, 
and  power.  DG  is  drawn  parallel  to  AB  and  QD  and  PF  are  per- 
pendicular to  AB. 

Ex.  9.  Construct  the  circle  diagram  for  a  25 ~  circuit  having 
a  constant  capacity  of  50  mf.,  a  constant  resistance  #2  =  10a,  and 
a  variable  resistance  rx  which  changes  from  (a)  5a  to  10a ;  (6) 
0  to  10a;  (c)  2.5  to  10a;  (d)  2.5  to  12.5a;  (e)  2.5  to  7.5a.  Deter- 
mine the  maximum  and  minimum  power  factors,  currents,  and 
power  for  the  elements,  as  well  as  for  the  entire  circuit. 

Observation.  In  a  circle,  diagram  any  line  which  is 
affected  by  a  single  variable  may  be  used  to  represent  the 
instantaneous  magnitude  of  that  variable.  The  chords  of 
a  circle  diagram  are  divided  into  segments  by  a  horizontal 
auxiliary  circle  when  their  resistance  or  reactance  elements  are 
constant  quantities.     The  chords  of  a  circle  diagram  are  divided 


THE  CIKCLE  DIAGRAM 


611 


into  segments  by  a  vertical  auxiliary  circle  when  their  resistance 
or  reactance  elements  consist  of  both  constant  and  variable 
quantities. 

8.  Multiple  Auxiliary  Circles.  In  Fig.  287  the  circuit 
consists  of  two  constant  and  one  variable  resistance  Rs, 
R2,  and  ri  respectively,  and  a  constant  reactance  X, 
which  are  supplied  with  current  from  a  source  of  constant 
E.M.F.  E. 

The  resistance  chord  AF  is  divided  into  the  segments 
AN  =  i(Ra+R2)  and  NF  =  ir\.  In  order  to  separate  the 
drops  across  R%  and  R2,  AN  is  subdivided  at  D  by  the 
second  vertical   auxiliary  circle  ADT\,    D  is  located  by 


Fig.  287. 


dividing  AN  into  two  segments  AD  and  DN  whose  ratio 
equals  Rs  :  R2.  DT  is  drawn  perpendicular  to  AD  and  a 
semicircle  is  drawn  through  A,  D,  and  T. 


(18) 


(19) 


AD^AN  =  DN  =  BF 
AT    AM     TM    AB} 


AD     constant  AT 


DN     constant  TM 


=  constant. 


Therefore  the  two  segments  AD  and  DN  of  the  entire 
chord  AF  which  are  intercepted  by  the  auxiliary  circles 
preserve  a  constant  ratio  #3  :  R2.  This  method  may  be 
extended  to  a  further  subdivision  of  the  constant  part  of 
the  total  resistance.     NB  represents  the  drop  across  the 


612 


PRACTICAL  MATHEMATICS 


impedance  z\  —  \/r\2-\-X2.  The  phase  angle  <$>  for  the  entire 
circuit  and  the  phase  angle  cjn  for  the  impedance  z\  may 
be  measured  from  the  perpendiculars  drawn  to  AB  and  NB 
respectively  as  indicated  in  Fig.  287.  Acute  angles  are 
equal  when  their  respective  sides  are  perpendicular. 

Ex.    10.    Fig.    288    represents   an   alternator   with    constant 
excitation  and  variable  non-inductive  load.    The  resistance  and 


Fig.  288. 

reactance  of  the  alternator's  armature  are  assumed  constant. 
Show  how  the  changes  of  the  terminal  voltage  may  be  read  from 
a  circle  diagram  as  the  load  resistance  varies. 

Ex.    11.     Construct   and   explain  the   diagrams   in   Fig.   289, 
which  represent  a  circuit  with  two  constant  reactances  Xi  and  X2 


Fig.  289. 


in  series  with  a  variable  resistance  n.    The  impressed  voltage  E 
is  constant. 

Ex.   12.     Construct  and  explain   the  diagrams  in  Fig.  289, 
which  represent  a  circuit  with  a  constant  reactance  X2  in  series 


THE  CIRCLE  DIAGRAM 


613 


with  a  constant  resistance  Ri  and  a  variable  reactance  xx.     The 
impressed  voltage  E  is  constant. 

Ex.  13.     Construct  and  explain  the  circle  diagram  in  Fig.  290, 
which  represents  a  circuit  with  variable  resistance  rx  in  series 


Fig.  290. 

with  a  constant  capacity  reactance  X2  and  a  constant  inductive 
reactance  Xx  in  which  X2>Xi.  The  impressed  voltage  E  is 
constant. 

Ex.  14.     Fig.  291  represents  the  diagram  for  a  series  circuit 


Fig.  291. 


consisting  of  two  impedances  2i=vriH^i2  and  z2  =  \Zr22-\-x22 
supplied  with  current  from  a  source  of  a  constant  E.M.F.  E. 
Why  does  DF  represent  the  •  energy  component  of  E  and  why 
does  FG  represent  the  wattless  component  of  E?  What  signifi- 
cance do  we  attach  to  lines  ST,  TF,  and  SF?  Can  a  single  line 
be  drawn  in  the  figure  to  represent  the  drop  across  Zi?  What  is 
the  difficulty  in  the  use  of  this  diagram? 

Ex.   15.     Construct  and  explain  the  use  of  the  circle  diagram 
to  correspond  to  the  circuit  shown  in  Fig.  292. 


Fig.  292. 


614 


PRACTICAL  MATHEMATICS 


Ex.    16.     Construct   and   explain   the   diagrams   in   Fig.    293 
which    represent    a    circuit    in    which    a     constant    impedance 


^g^ 


R,  x, 


ra  x2 


Fig.  293. 

Zi=  \/Ri2-\-Xi2  is  in  series  with  a  variable  impedance  Z2=\/r22+X2. 
Construct  a  second  circle  diagram  in  which  the  chord  AG  is  divided 
by  an  eccentric  auxiliary  circle. 

Xi 


Fig.  294. 
9.  Reactance  and  Resistance  in  Series  with  a  Con- 
stant Voltage  Circuit.  In  practice  it  becomes  necessary 
to  maintain  a  constant  voltage  E  across  an  impedance. 
If  an  additional  resistance  or  reactance  is  placed  in  series 
with  the  impedance,  then  the  total  generator  voltage  e  to 


THE  CIRCLE  DIAGRAM 


615 


be  applied  to  the  circuit  may  be  determined  by  the  circuit 
diagram  as  illustrated  in  Figs.  294  and  295. 

In  both  case  A  B  =  E\s  the  diameter  of  a  circle  diagram  and  is 
the  drop  across  the  impedance.  AF  in  Fig.  294  is  an  extension 
of  AD  and  is  the  additional  voltage  drop  across  the  additional 


resistance^.  FB  =  e  is  the  generator  volt  age.  BjF  in  Fig.  287 
is  an  extension  of  DB  and  is  the  additional  voltage  drop  across 
the  additional  reactance  X2.  AF  =  e\s  the  generator  voltage. 
Fig.  296  shows  a  modification  of  Fig.  287  for  the  case 
in  which  the  extra  reactance  £2  is  variable.     Explain  Fig.  287. 


616 


PRACTICAL  MATHEMATICS 


10.  Parallel    Circuits   with    Constant  Applied  Voltage. 
Fig.  297  represents  a  circuit  with  two  parallel  branches  I 

and  II  which  are  both  supplied  by 
the  same  constant  voltage.  Branch 
I  contains  a  constant  resistance  R± 
and  a  variable  reactance  x\. 
Branch  II  contains  a  constant 
resistance  R2  and  a  variable  re- 
actance X2.  Each  branch  has  its 
distinct  circle  diagram  as  repre- 
sented by  (a)  and  (6)  respectively 
in  Fig.  298.  The  current  in  I  is 
proportional  to  AF  =  i\R\  and  the 
current  in  II  is  proportional  to  DB  =  i2R2.  In  order 
to  obtain  the  resultant  current  for  the  entire  circuit, 
the  circle  diagrams  for  I  and  II  must  be  superposed  and 
modified  so  as  to  place  the  vectors  AF  and  DB  adjacent 
and  consecutive  and  they  must  both  be  represented  to  the 


Fig.  297. 


Fig.  298. 

same  scale.  The  first  requirement  is  satisfied  by  inverting 
(6)  into  (c)  of  Fig.  298.  The  vectors  DB  and  BF  remain 
parallel  and  equal  in  their  new  positions  FH  and  GF 
respectively,  and  in  the  new  position  (c)  FH  lags  behind 
E  in  accordance  with  the  lagging  of  DB  behind  E  in  (b). 
If  H  were  now  placed  in  coincidence  with  A,  the  vectors 


THE  CIRCLE  DIAGRAM  617 

AF  and  FH  would  be  consecutive.  The  second  requirement 
is  satisfied  by  changing  the  diameter  of  circle  (c)  as  shown 
in  (d)  of  Fig.  298  according  to  (24).  In  (a),  (6),  and  (c) 
the  voltage  lines  AF,  FC,  DB,  BF,  GF,  FH  are  all  measured 
to  the  same  scale  as  the  three  equal  diameters  AC  =  DF 
=  GH  =  E.  When  AF  measures  the  current  i\  in  I  then 
the  scale  of  i\  is  expressed  in  (20) . 

(20)  fc-fe, 

/01  x  i      r  .      scale  of  Erx     scale  of  E 

(21)  scale  of  ti  = = = 5 . 

When  DB  measures  the  current  %2  in  II  then  the  scale 
of  %2  is  expressed  in  (23) . 

(22)  i2  =  ^ 


fli' 


/rvox  ,      r  .      scale  of  Er2     scale  of  E 

(23)  scale  of  %2  = 


(24) 


R2  R2 

scale  of  i\  _R2 
scale  of  \2    Ri 


(24)  is  obtained  from  dividing  (21)  by  (23)  and  states 
that  the  scales  of  the  current  lines  in  (a)  compared  with 
(6)  or  (c)  are  inversely  proportional  to  the  constant  resistances 
of  those  respective  branches.  In  order  to  change  the  scale 
of  i*2  so  as  to  correspond  to  the  scale  of  ti  it  is  necessary 

to  change  the  length  of  12  by  the  ratio  of  — .     This  is  accom- 

R\ 

plished  most  easily  by  changing  the  diameter  of  the  circle 

#2 

(6)  or  (c)  in  the  ratio  —  as   shown  in    (d)    which   has  the 
Ri 

effect  of  changing  all  lines  of  (d)  in  the  same  ratio.     This 

ratio  is  called  the  diametric  scale  factor. 

,  FH      GF  =GH  =R2 

^    }  FlH    GlFl     GlH    R{ 


618  PEACTICAL  MATHEMATICS 

Instead  of  changing  the  diameter  of  circles  (6)  or  (c) 
the  diameter  of  circle  (a)  may  be  changed  by  multiplying 

r> 

it  by  a  corresponding  diametric  scale  factor  — .     Fig.  299 

represents  the  circle  diagrams  for  the  two  branch  circuits 

7? 

I  and  II  described  above.     The  diametric  scale  factor  — 

R2 
is  applied  to  the  circle  diagram  for  branch  I  and  therefore 

the  length  of  CF  equals  —  times  the  length  of  AB. 
H2 

The  vectors  AC  and  CD  representing  i\  and  1*2,   are 

consecutive  and  constructed  to  the  same  scale,  therefore 


^("rD 


Fig.  299. 

their  resultant  i  is  AD  and  is  also  measured  with  a  like 
scale.  Since  the  phase  angles  §\  and  §2  for  i\  and  12 
respectively  are  measured  from  horizontal  reference  lines 
then  the  phase  angle  cj>  for  the  resultant  i  is  also  measured 
from  the  horizontal  reference  line. 

11.  Another  alternative  method  for  the  above  problem 
is  to  use  a  single  circle  for  both  circuits  I  and  II  as  shown 
in  Fig.  300.  Its  principal  disadvantage  lies  in  the  fact  that 
more  construction  lines  are  necessary  and  there  is  no  con- 
venient method  for  applying  the  unequal  scales  to  the  two 
lines  AC  and  AF  in  order  to  determine  AP  and  AM 
respectively. 


THE  CIRCLE  DIAGRAM 


619 


Fig.  300. 

Another  modification   of  the   above  problem  is  shown 
in  Fig.  301.     The  second  circle  diagram  CDB  need  not  be 


Fig.  301. 

transposed  if  FP  is  drawn  parallel  and  equal  to  DB.     Then 
the  resultant  AP  is  the  vector  sum  of  AF  and  FP. 

12.  Relabeling  the  Lines  of  a  Circle  Diagram.  Fig. 
302  shows  the  circle  diagram  for  a  branch  circuit.  The 
circle  for  circuit  I  is  drawn  with  a  diameter  CF  =  E 
and  therefore  the  current  i\  is  proportional  to  CD.     For 

circuit  II  the  diameter  AC  =  —  E  and  therefore 

#2 
/l2 


620 


PRACTICAL  MATHEMATICS 


The  three  lines,  CD,  BC,  and  BD  are  proportional  to  %it  1*2, 
and  i  and  therefore  CD  =  i\R\}  BC  =  %2Ri,  and  BD=iRi. 

£C2 


Fig.  302. 

Ex.  17.     In  Fig.  302  show  that  AB  =i2X2  -^. 

13.  The  Determination  of    the    Equivalent   Resistance 
and  Equivalent  Reactance.     Fig.  303  represents  the  circle 


Fig.  303. 

diagram  for  a  parallel  circuit  having  a  constant  resistance 
and  a  variable  reactance  in  each  branch.  The  total  current 
i  is  represented  by  FD  =  iR\  which  has  the  same  scale  as 
FB  and  BD.    The  total   current  lags   cj>  degrees  behind 


FD 

iRi     Ri 

BD'~ 

iRa        Ra 

Ra- 

D'C 
BD'~~ 

IXa       Xa 
lRa      Ra 

THE  CIRCLE  DIAGRAM  621 

the  E.M.F.  and  therefore  BD'  may  be  drawn  parallel  to 
FD  so  as  to  represent  the  total  current  in  a  circle  whose 
diameter  is  E.  If  we  designate  the  equivalent  resistance 
and  equivalent  reactance  of  the  whole  circuit  by  Ra  and  x0} 
then  BD'  =  iRa,  D'C  =  ixa,  and  BC  =  E  =  iza. 


(26) 
(27) 
(28) 
(29)  Xa  =  RaX^±f  =  RlX^ 

Interpret  (27)  and  (29). 

Ex.  18.  Construct  the  circle  diagram  for  a  two-branch  circuit 
which  is  supplied  with  a  constant  E.M.F.  E.  Branch  I  contains 
a  constant  resistance  #1  =  1  Oft  and  a  variable  reactance  X\,  ranging 
from  5ft  to  7.5ft.  Branch  II  contains  a  resistance  R2  equal  to  7.5ft 
and  a  variable  resistance  x2,  ranging  from  5ft  to  7.5ft.  Designate  the 
current  lines,  power  lines,  and  power  factor  lines  for  each  branch 
and  for  the  entire  circuit.  Determine  the  equivalent  resistance 
of  the  circuit  and  the  equivalent  reactance  when  Xi  and  x2  reach 
maximum  and  minimum  values  simultaneously.     See  Fig.  315. 

Ex.  19.  Construct  the  circle  diagram  for  the  circuit  described 
in  Ex.  18,  excepting  that  the  resistances  are  variable  and  the 
reactances  are  constant.  Interchange  the  values  for  R  and  x 
given  in  Ex.  18.     See  Fig.  316. 

14.  Parallel  Circuits  Whose  Constant  Elements  are 
Unlike.  Fig.  304  represents  a  circuit  with  two  parallel 
branches  supplied  with  a  constant  E.M.F.  E.  Branch  I 
contains  a  constant  resistance  Ri  and  a  variable  reactance 
x\.  Branch  II  contains  a  constant  reactance  X2  and  a 
variable  resistance  r2.  In  circuit  I  the  current  is  pro- 
portional to  Ei,  whereas  in  circuit  II  the  current  is  pro- 


622 


PRACTICAL  MATHEMATICS 


portional   to   EXi.     The    diameter   of  the   circle   diagram 

for  circuit  II  will  be  obtained   by  multiplying  E  by  — - 

X2 

which  is  the  diameter  scale  factor. 


(30) 
(31) 

(32) 


CA=i2X2X^=i2Ri, 

A  2 

CF=i2r2¥2- 

The  diameter  of  semicircle  II      Ri 
The  diameter  of  semicircle  I      X2 


Fig.  304. 


CD  is  the  vector  sum  of  the  currents  i\  and  12  since  the 
three  lines  CD  =  iRi,  CA=%2R\,  AD  =  iiRi,  are  proportional 
to  i,  i\  and  12.  AG  is  drawn  parallel  to  CD  and  therefore 
AG  is  the  drop  across  the  equivalent  simultaneous  resist- 
ance and  GE  is  the  drop  across  the  equivalent  simultaneous 
reactance.     Do  Eqs.  (27)  and  (29)  apply  to  this  circuit? 


THE  CIRCLE  DIAGRAM 


623 


Ex.  20.  Construct  and  explain  the  diagram  in  Fig.  305  which 
applies  to  a  two-branch  circuit  supplied  with  a  constant  E.M.F.  E. 
In  branch  I  there  is  a  constant  reactance  Xx  and  a  variable  resist- 


MA/W^/WVWNQflflOOOQOOO. 


ance  n.  In  branch  II  there  is  a  constant  resistance  R2  and  a 
variable  reactance  x2.  Supply  any  necessary  lines  to  show  power 
factors  and  power  in  each  branch  and  for  the  total  circuit. 

15.  The  Circle  Diagram  for  a  Constant  Current  Circuit. 

The  circle  diagram  is  applied  to  simple  circuits  supplied 
with  constant  current.     The  diameter  of  the  semicircle  in 


Fig.  306  represents  the  constant  current  I  which  is  resolved 
into  two  orthogonal  components  %q  and  ib  the  locus  of  whose 


624 


PRACTICAL  MATHEMATICS 


intersection  is  the  semicircle.  For  this  circuit  the  con- 
ductance G  is  constant  and  the  susceptance  b  is  variable. 
Which  lines  are  proportional  to  the  applied  voltage,  power, 
power  factor,  and  impedance,  respectively. 

The  diameter  of  the  semicircle  in  Fig.  307  represents 
the  constant  current  I  which    is  resolved  into  i§  and  ib 


Fig.  307. 


the  locus  of  whose  intersection  is  the  semicircle.  For  this 
circuit  the  susceptance  B  is  constant  and  the  conductance 
g  is  variable.  Which  lines  are  proportional  to  the  applied 
voltage,  power,  power  factor,  and  impedance,  respectively. 

16.  Circuits  Having  Constant  Resistance  and  Con- 
stant Inductance  with  Variable  Frequency.  Fig.  308  is 
a  modification  of  the  circle  diagram  and  is  used  for  showing 
the  relation  between  the  current  and  slip  of  an  induction 
motor.  If  we  assume  a  constant  excitation  for  an  alternator, 
then  its  generated  E.M.F.  e  will  depend  upon  its  speed  s. 
The  total  reactance  x  of  the  circuit,  including  the  alternator, 
will  also  depend  upon  the  speed.  R  is  the  constant  resistance, 
L  the  constant  inductance,  and  i  the  current  of  the  circuit 
and  K  is  a  proportionality  factor. 


(33) 


eozs,     xccs, 


eozx. 


(34) 


Kx. 


The  E.M.F.  e  varies  directly  as  the  reactance  of  the 
circuit  within  the  limits  of  saturation  of  the  generator. 


THE  CIRCLE  DIAGRAM 


625 


(35) 


^  = 


Kx 


Vr2+x2   Vh2+x2' 


(36) 


ICC 


Vr2+. 


In  Fig.  308  A B  represents  the  maximum  value  of  x 
and  AD  the  diameter  of  the  semicircle  AKD  is  constructed 
equal  to  R.  Then  DB  represents  the  maximum  impedance 
of  the  circuit.     With  a  diminution  of  speed  or  frequency 


the  value  of  x  decreases  so  that  when  x=AF,  then 
FD  =  VR2+x2.  Join  A  with  K  the  intersection  of  FD  with 
the  semicircle.  Then  AKD  is  a  right  angle  inscribed  in  a 
semicircle. 


(37) 
but 

(38) 

(39) 
(40) 


AF=sAK  =  _x__ 
FD    AD     y/W+j? 

AK 
35= an*, 

'.     i  oc  sin  cj), 
.     iccAK. 


Therefore,  the  current  line  is  represented  by  AK.  'What 
lines  represent  the  power  factor,  the  total  volts  generated, 
and  the  active  power? 


626 


PEACTICAL  MATHEMATICS 


Ex.  21.     Construct  and  explain  the  diagrams  in  Fig.  309  in 
which  BF=EC=  ixc  and  DF=EL=  ixL. 


Ex.  22.     Construct  and  explain  the  diagram  in  Fig.  310  which 
lines  in  the  figure  represent  active  and  wattless  power. 


£C\_  OCq 


Fig.  310. 


Ex.  23.     Construct  and  explain  the  diagram  in  Fig.  311.    Add 

a  diagram  of  the  circuit. 


Fig.  311. 


THE  CIRCLE  DIAGRAM 


627 


17.  Multiple  Circuit  Circle  Diagrams.  Fig.  312  repre- 
sents a  multiple  constant  voltage  circuit  with  three  branches 
each  containing  a  resistance  and  an  inductive  reactance. 


Fig.  312. 


In  circuits  I  and  III  the  reactances  are  constant  and  the 
resistances  are  variable  whereas  in  II  the  resistance  is 
constant    and    the  reactance    is    variable.     Fig.    313    was 


Fig.  313. 

constructed  by  first  representing  the  circle  diagram  FGH 
for  circuit  II,  in  which  FG  represents  %2.     The  diametric 

scale  factor  —  is  applied  in  constructing  the  circle  diagram 
A3 


628 


PRACTICAL  MATHEMATICS 


DCF  for  circuit  III  in  which  CF  represents  ^3.     The  diametric 
■p 

scale  factor  — -  is  applied  in  constructing  the  circle  diagram 
Xi 

ABC  for  circuit  I  in  which  BC  represents  i\.     The  phase 

angles  for  ii,  12,  13,  and  i  are  measured  from  horizontal 

reference   lines.     The   total   current   i  =  BG   is   the   vector 

sum  of  BC,  CF,  and  FG. 

Ex.  24.    Fig.  314  represents  a  multiple  circuit  in  which  the 
three  branches  I,  II,  and  III  each  contain  resistance,  inductance, 


Fig. '314. 


and  capacity.  In  I  both  resistance  n  and  capacity  reactance  xc\ 
are  variable.  In  II  the  resistance  r2  and  the  inductive  reactance 
XL2  are  variable.  In  circuit  III  the  capacity  reactance  xc3  only 
is  variable.  Discuss  the  diagram  for  the  possible  conditions  when 
one  of  the  two  variables  in  each  circuit  reduces  to  zero. 


THE  CIRCLE  DIAGRAM 


629 


TABLE  XXXVI. RESISTANCE  OF  SOFT  OR  ANNEALED 

COPPER  WIRE 


Diameter 

Area  in 

Ohms 

Diameter 

Area  in 

Ohms 

B.  &S. 

Gauge. 

in  Mil3. 
d. 

Circular 
Mils. 

per 
1000  ft. 
at  20°  C. 
or  68°  F. 

B.  &S. 
Gauge. 

in  Mils, 
d. 

Circular 

Mils. 

d*. 

per 

1000  ft. 

at  20°  C. 

or  68°  F. 

0000 

460.00 

211,600 

.04893 

19 

35.890 

1288.1 

8.038 

000 

409.64 

167,810 

.06170 

20 

31.961 

1021.5 

10.14 

00 

364.80 

133,080 

.07780 

21 

28.462 

810.10 

12.78 

0 

324.86 

105,530 

.09811 

22 

25.347 

642.40 

16.12 

1 

289.30 

83,694 

.1237 

23 

22.571 

509.45 

20.32 

2 

257.63 

66,373 

.1560 

24 

20.100 

404.01 

25.63 

3 

229.42 

52,634 

.1967 

25 

17.900 

320.40 

32.31 

4 

204.31 

41,742 

.2480 

26 

15.940 

254.10 

40.75 

5 

181.94 

33,102 

.3128 

27 

14.195 

201.50 

51.38 

6 

162.02 

26,250 

.3944 

28 

12.641 

159.79 

64.70 

7 

144.28 

20,816 

.4973 

29 

11.257 

126.72 

81.70 

8 

129.49 

16,509 

.6271 

30 

10.025 

100.50 

103.0 

9 

114.43 

13,094 

.7908 

31 

8.928 

79.70 

129.9 

10 

101.89 

10,381 

.9972 

32 

7.950 

63.21 

163.8 

11 

90.742 

8,234.0 

1.257 

33 

7.080 

50.13 

206.6 

12 

80.808 

6,529.9 

1.586 

34 

6.305 

39.75 

260.5 

13 

71.961 

5,178.4 

1.999 

35 

5.615 

31.52 

328.4 

14 

64.084 

4,106.8 

2.521 

36 

5.000 

25.00 

414.2 

15 

57.068 

3,256.7 

3.179 

37 

4.453 

19.82 

522.2 

16 

50.820 

2,582.9 

4.009 

38 

3.965 

15.72 

658.5 

17 

45.257 

2,048.2 

5.055 

39 

3.531 

12.47 

830.4 

18 

40.303 

1,624.3  6.374 

40 

3.145 

9.89 

1047 

630 


PEACTICAL  MATHEMATICS 


Fig.  315. 


Fig.  316. 


APPENDIX 


GREEK  ALPHABET 


Letters. 

Name. 

Letters. 

Name. 

A  a 

Alpha 

N  v 

Nu 

B  p 

Beta 

B  £ 

Xi 

ry 

Gamma 

O  o 

Omicron 

A  8 

Delta 

IT    TV 

Pi 

E    € 

Epsilon 

Pp 

Rho 

z  t 

Zeta 

%    O"    S 

Sigma 

H  rj 

Eta 

T  r 

Tau 

0  0 

Theta 

Y  v    ■ 

Upsilon 

I     i 

Iota 

<&  <j> 

Phi 

K     K 

Kappa 

x  X 

Chi 

A  X 

Lambda 

*  I 

Psi 

M  fx 

Mu 

12   o) 

Omega 

INSTRUCTIONS  FOR  PREPARING  AND  REPORTING  DAILY 

WORK 

1.  The  Work-book.  The  student  should  provide  him- 
self with  a  Mathematics  Work-book,  which  consists  of  a 
work-book  cover  intended  to  hold  300  sheets  of  20-pound 
plain  linen  paper,  measuring  5|X8J  ins.  The  work-book 
cover  has  a  flexible  leather  back  and  is  perforated  with  two 
pairs  of  eyelets.  Attach  20  sheets  of  the  plain  linen  paper 
to  the  rear  cover  for  use.     The  front  cover  should  be  used 

631 


632  PKACTICAL  MATHEMATICS 

for    finished    work.    Procure    a   good   fountain    pen    and 
fountain-pen  ink. 

2.  Identity  of  Property.  Print  a  suitable  label  as 
directed  and  place  it  on  the  outside  of  the  front  cover  of 
the  work-book.  Enter  your  name,  class,  home  and  board- 
ing address  on  the  inside  of  the  front  cover  of  both  your 
text-book  and  work-book.  Take  sufficient  precaution  to 
mark  every  piece  of  your  personal  property  so  that  it  may 
be  distinguished  and  identified  easily. 

3.  The  Text.  The  work-book  is  not  to  be  used  for 
notes  as  the  text  is  deemed  ample,  but  should  any  necessity 
arise  for  adding  additional  information,  it  may  be  entered 
in  the  text-book  in  the  blank  space  at  the  end  of  a  chapter. 
Both  text-book  and  work-book  should  be  carried  daily 
to  the  class  room  unless  instruction  is  given  to  the  contrary. 

4.  Preparation  of  Daily  Work.  All  work  should  repre- 
sent individual  effort  and  should  be  done  neatly  and 
accurately  in  black  writing  ink.  Drawings  and  diagrams 
should  be  constructed  lightly  with  a  hard  pencil  and  straight- 
edge and  after  the  instructor  has  given  his  approval  they 
should  be  inked  with  India  ink.  Title  pages  should  be 
lettered  in  India  ink  on  the  work-book  paper,  and  should 
precede  each  important  division  or  chapter  of  the  text. 
The  main  title  page  should  be  protected  by  a  blank  fly- 
leaf. Titles  should  be  brief  and  lettered  across  the  center 
of  the  page.  Before  entering  the  class  room  place  your 
name  on  each  page  at  the  top  between  the  holes.  Place 
the  date  when  the  work  is  due  at  the  upper  right-hand 
corner  one  inch  from  the  top  of  the  paper.  As  the  finished 
pages  accumulate  enter  consecutive  page  numbers  in  red 
ink  at  the  lower  right-hand  corner  of  the  page.  One  inch 
from  the  upper  left  corner  of  the  page  write  a  fraction 
whose  numerator  is  the  text  page  number,  and  whose 
denominator  is  the  exercise  number. 

All  work  should  be  written  horizontally  across  the 
page  and  if  the  student  experiences  any  difficulty  from 


APPENDIX  633 

careless  previous  training  he  should  procure  or  prepare 
a  ruled  sheet  of  paper  with  guide  lines  spaced  one-quarter 
of  an  inch  apart.  Three  vertical  lines  on  the  ruled  sheet 
will  prove  of  assistance.  Headings  should  be  carefully 
printed,  centered  and  spaced.  The  work  should  be  legible 
and  carefully  spaced,  not  more  than  one  equation  should 
appear  on  a  line.  Incidental  calculations  should  appear 
on  the  paper  near  the  right  margin.  Special  care  should 
be  given  to  the  formation  of  numerals,  and  symbols  and 
more  especially  to  the  equality  sign  and  the  parenthesis. 
A  five-minute  daily  practice  making  alphabetic  characters 
and  symbols  will  prove  very  profitable.  Wherever  con- 
venient arrange  the  steps  of  your  work  so  that  equality 
symbols  are  placed  in  a  vertical  column.  All  proofs  of 
analytic  statements  must  bear  an  equation  number  placed 
in  a  parenthesis  at  the  left  of  the  line  and  also  a  mathematic 
authority  placed  at  the  right  of  the  line.  All  equations 
must  be  numbered  in  sequence  for  each  exercise. 

All  incorrect  or  rejected  work  must  be  presented  again 
showing  the  corrections  in  red  ink,  at  or  before  the 
beginning  of  the  period  immediately  following  its  non- 
acceptance. 

An  excuse  for  the  non-performance  of  any  assigned 
work  must  be  written  upon  the  regular  work  paper  dated 
and  signed  and  presented  in  lieu  of  the  work. 

5.  Presentation  of  Work.  On  the  stroke  of  the  bell  at 
the  beginning  of  the  period  all  assigned  work  for  the  day, 
excuses  and  delinquent  work,  must  be  passed  in  each  row 
so  as  to  be  received  in  alphabetic  order  by  the  end  man 
in  each  row.  The  rear  end  man  will  collect  all  work  and 
deposit  it  in  a  designated  drawer  of  the  instructor's  desk. 
In  order  that  this  may  be  done  quietly  and  without  delay 
all  work  should  be  ready  for  collection  when  the  student 
enters  the  class  room,  and  he  should  make  an  immediate 
entry  on  the  daily  record  sheet  which  has  been  distributed 
by  the  end  man  in  his  row. 


634  PRACTICAL  MATHEMATICS 

6.  Reporting    Daily    Work.    A   daily   record    sheet   is 

attached  to  the  front  of  the  student's  accepted  work,  and 
contains  sufficient  space  for  twenty  daily  entries  under 
the  following  column  headings:  date,  pages  and  exercises, 
title  of  matter  assigned  for  preparation;  a  sub-divided 
time  column  for  indicating  daily  the  number  of  hours  spent 
for  each  prepared  subject  in  the  course;  two  other  columns 
headed  "  foreman  "  and  "  remarks  "  are  reserved  for  the 
instructor's  use. 

When  the  time  sheet  is  filled  and  the  work  described 
thereon  has  been  accepted,  then  the  work  and  the  time 
sheet  is  placed  in  charge  of  a  custodian,  who  stores,  arranges 
and  credits  it  upon  a  designated  shelf  in  the  class  room 
closet,  from  which  it  may  be  removed  only  upon  special 
permission  from  the  instructor. 


INDEX 


Abbreviation  by  symbols,  6,  256 

Abscissa,  283 

Absolute  term,  247,  310 

Abvolt,  193 

Acceleration,  476 

Addition,  arithmetic  4;  algebraic  7;  impedances  of  curves,  323;  vectors, 
526 

Admittance,  545 

Algebraic  equation,  317 

Alternating  current  problems,  573 

Altitude,  49 

Ampere-hour,  133 

Ampere-turns,  184,  185 

Amplitude,  408,  412 

Analysis  of  formulas,  126  , 

Angle,  acute,  adjacent,  alternate,  alternate-exterior,  alternate-interior, 
central,  complementary,  direct,  exterior,  exterior-interior,  inscribed 
interior,  obtuse,  right,  straight,  supplementary,  supplementary- 
adjacent,  unilateral,  vertical,  23,  26-31;  bisection  of,  30,  373; 
functions  of,  396,  397;  measurement  of,  29,  393;  multisection  of, 
380;   radian  of,  393;    theorems  on,  35,  36,  37;    trisection  of,  377 

Angular  velocity,  401,  404 

Annulus,  area  of,  205 

Antecedent,  53 

Antidifferentia^ion,  447;  graphic^  455 

Antilogarization,  99 

Antioperation  447 

Apex,  60 

635 


636  INDEX 

Approximations,  determination  of  error  of,  240,  241 

Arc,  degree  of,  28,  29;   length  of,  206;   of  sine  curve,  404;   radian  of, 

29,  393;  rectification  of,  206. 
Arches  of  sine  curve,  404 
Area,  46,  203;  center  of ,  264;  circle,  204;  circular  sector  and  segment, 

209;  ellipse,  205;  elliptic,  hyperbolic  and  parabolic  segment,  210, 

211;     parallelogram,    203;     quadrilateral,    51;     trapezoid,    204; 

triangle,  51,  203;  regular  polygon,  204;  unit  of,  46 
Area  of  irregular  figures,  by  Durand's  rule,  mean  ordinate  rule,  mid- 

ordinate  rule,  panimeter,  Simpson's  one-third  and  three-eighths 

rules,  squared  paper,  trapezoidal  rule,  Weddle's  rule,  weighing, 

211-214 
Area,  elementary,  464;    indicator  diagram,   241;    ship   section,   241; 

sine  loop,  467;  sin2  curve,  463 
Areal  formula  for  integration,  464 
Armature  bearing,  276 
Arrow-head,  525 

Authorities  in  mathematics  work,  35 

Auxiliary  semicircle,  eccentric,  horizontal,  multiple,  vertical,  605-608 
Average  E.M.F.,  193;  value  sine  loop,  467 
Axiom,  15,  31,  32,  33,    addition  32,  38,  39;    construction  35,  38,  39; 

direction,  27,  35;   division,  32;   equality,  15,  32,  38,  40;   fractions, 

76,  78;  magnitude,  27,  35,  38,  40;  motion,  35;  multiplication,  32; 

operations,  31;  plane,  35;   point-line,  35;   power,  32;   reductio  ad 

absurdum,  35;  root,  32;  subtraction,  32 
Axal  line,  367 
Axis,  neutral,  268;    of  abscissas,  297;    of  coordinates,  287;    ordinates, 

297;  reference,  280,  282;  transformation  of,  286 


B 

Back  pressure,  176 

Balanced  E.M.F.,  in  resistive-reactive  circuits,  540,  543 

Base,  49,  95,  203,  218;  logarithmic  common  and  Napierian,  102 

Beam,  261;  cantilever,  263;  I- and  T-,  271;  deflection  of,  273;  stiffness 

of,  273;  strength  of,  271,  272 
Binomial,  82;  theorem,  240 
Biquadractic  inches  268 

Bisection  of  a  line,  31,  36;    an  angle,  30,  38,  373;    parallelogram,  49 
Branches  of  curves,  324 
Brigg's  system  of  logarithms,  102 
British  thermal  unit,  166 


INDEX  637 


Calorie,  pound,  166;  small,  167 

Capacity,  511-521;  formulas  for,  516;  of  condensers,  513;  practical 
values  of,  515;  reactance,  522;  carrying,  of  wires,  298 

Causes,  53 

Center  of  area,  gravity  mass,  264;  of  circle,  27 

Centroid,  264 

Ceradini's  arc  rectification,  206 

Change  in  place,  time,  stress,  279 

Channel,  271 

Characteristic,  96;  curves,  483;  compound  generator,  485;  efficiency 
shunt  motor,  489;  load  and  voltage  shunt  motor,  483;  series 
generator,  487;  starting  torque  shunt  and  series  motors,  490; 
storage  battery,  488 

Chart,  279;    of  charts,  391;    of  lines,  383;  reciprocal,  388 

Charting  of  linear  formulas,  389;  reciprocal  formulas,  388 ;  charting  of 
three  or  more  variables,  388-392 

Chords,  28;  table  of,  121 

Chordel,  381 

Circle,  area  of,  204;  auxiliary,  605;  center  of,  27;  diagrams,  601 ;  eccen- 
tric, 609;  equation,  332;  polar  equation,  371;  section  of  cone,  330 

Circuits  having  resistance,  inductance  and  capacity,  543 

Circular  arc,  length,  206;  mil  foot,  138,  202;  rectification,  206;  ring, 
205;  sector  and  segment,  209;  unit,  202 

Circularization  of  a  line,  208 

Circumference,  28 

Coefficient,  7;  of  elasticity,  260;  of  time,  406 

Cofunction,  66 

Coil,  field  within  a,  184 

Coincidence,  40 

Common  denominator,  79 

Components,  curves,  419;  of  an  impressed  E.M.F.,  in  an  inductive 
reactive  circuit,  541 

Composition  of  curves,  323 

Compound,  magnetic  circuit,  185;  curve,  423;  fraction,  81;  interest 
law,  363 

Compression,  259 

Concentric  circle,  29 

Conehoid,  377 

Conclusion,  39 

Concurrent  forces,  532 


638  INDEX 

Conductance,  163 
Conductivity,  183 
Cone,  60,  61;    frustum  of,  231;    generation  of,  234;    lateral  surface, 

233;  right  circular,  330;  volume  of,  229,236 
Conic  sections,  330;   test,  330;   eccentricity  of,  342;  properties  of,  347 
Consequent,  53 
Construction  lines,  39 
Constant,  51 
Condensers,  512-521 
Continued  equality,  52 
Contraction,  396 
Convention  of  signs,  68 
Conversion  formulas,  398 

Coordinates,  linear,  383;  of  a  point,  297;  polar,  367 
Copper  loss,  179 
Corona  effect,  363 
Cosecant,  65 
Cosine,  64;  curve,  294;  hyperbolic,  366,  399;  law  of,  91;    series,  398; 

table  of,  121 
Cotangent,  65;  table  of,  121 
Coulomb,  133,  513 
Counterclockwise  direction,  64 
Counter  E.M.F.,  169,  176,  505 
Courses,  72 
Crest,  327 

Critical  frequency,  547 
Cross-section  paper,  logarithmic,  351,  357;  semilog,  358;  special,  359- 

362;  squared,  22,  280 
Cube,  218;    volume  of,  220;    decomposition,  228 
Curve,  280,  284;  branches,  324;  composition,  323;  damped  oscillation, 

427;     hyperbolic,    325;     logarithmic,    292;     magnetization,    288; 

parabolic,  317;   periodic,  294;    product  of  two;   423,  425;    scale 

of  integral,   461;    symmetrical,   323;    writing  equation  of,   348 
Cutting  plane,  330 
Cycle,  404,  547 
Cycloid,  217,  314,  377 
Cylinder,  generation  of,  234;    oblique,  226;    right,  226;    surface  of, 

233;  volume,  227,  236 


Damped  or  decaying  oscillation,  427 
Data,  75,  124,  279;  plotting,  280 


INDEX  639 

Decimal,  point,  1,  19;  system,  2 

Definite  integral,  465 

Deflection  of,  beam,  273;  instrument,  503 

Degree  of,  accuracy,  21,  22;    angle,  29;    arc,  29;    equation,  247,  306; 

integral  equation,  461;  term,  306 
Demonstration,  35-40 
Denominator,  20;  least  common,  79 
Departure  of  line,  525 
Derivative,  430;    of  a  power,  442,  444;    space,  435;    second,  459; 

table  of,  450-454 
Development  of,  cone,  61,  233;  cylinder,  59,  233 
Diagonal,  43;  prism,  221 
Diagram  of  lines,  383 
Diagrammatic  setting,  109,  110 
Diameter,  pulley,  276;    shaft,  275;    wire  for  power,    177;    wire  for 

winding,  199 
Die,  57,  58 

Dielectric  constant,  514 

Differentiation,  of  formulas,  442;  graphic,  439 
Differential  coefficient,  435;  table  of,  450-454 
Digit,  1 

Dimensions,  49 

Direct,  setting,  110;  variation,  52 
Direction,  counterclockwise,  64;  of  vector,  525 
Directral  distance,  342 
Directrix,  342 
Dividend,  20 
Division,. algebraic,  85,  86;  arithmetic,  20;  contracted,  20;  logarthmic, 

100;  slide  rule,  111,  112 
Dodecahedron,  239 
Double  integration,  459 
Durand's  rule,  216 

E 

Eccentric  circle,  609 

Eccentricity  of  conies,  342 

Eddy  current  loss,  192 

Effects,  53 

Effective  value,  current,  468;  mean,  463 

Efficiency,  commercial,  179;  electrical,  179;  generator,  168,  179; 
gross,  180;  joint,  277;  motor,  168,  177,  179;  net,  179;  trans- 
formation, 167;  tramsmission,  167 

Elasticity,  modulus  of,  260 


640  INDEX 

Elastic  limit,  260 

Electric,  instruments,  495;  pair,  511;  power,  165 

Elementary  area  or  strip,  464,  471 

Elements,  geometric,  60;  of  equation,  247;  of  graph,  348 

Ellipse,  area  of,  205,  eccentricity,  342;  equation,  332;  properties,  347; 
section  of  cone,  330 

Ellipsoid,  234 

Energy,  equation  of  A.  C.  circuit,  521;  loss  due  to  hysteresis,  191; 
loss  due  to  eddy  currents,  192 

Epicycloid,  375 

Equal,  41;  figures,  40 

Equality  axiom,  15 

Equation,  degree,  306;  degree  of  integral,  461;  homogeneous,  256; 
linear,  300-315;  non-linear,  316-341;  non-linear  simultaneous, 
336;  polar  of  a  circle,  371;  quadratic,  247;  simple,  247,  simul- 
taneous, 128;  writing  the,  of  a  curve,  348 

Equiangular  spiral,  372 

Equivalent,  impedance,  559,  reactance,  620;  resistance,  620 

Error,  area-linear,  239;  linear-square,  239;  linear- volume,  240;  mixed, 
240;  of  approximations,  240 

Evaluation,  17,  465 

Even-odd,  hyperbolas  and  parabolas,  320 

Evolute,  377 

Excess,  9,  10,  544 

Expansion,  396 

Exponent,  15,  16,  95,  247 

Exponential,  363;  family,  364,  365;  series,  398;  table,  366 

Expression  for  an  A.  C.  circuit,  544 

Extrapolation,  284,  352 

Extremes,  53 


Factor,  14,  508;  highest  common,  87 

Factorial  symbols,  240 

Factoring,    binominals,    89;     grouping,    89;     polynomials,    89;     tri- 

nominals,  89 
Families  of  curves,  320 
Farad,  513 

Field  within  a  coil,  184 
Figure,  center  of,  264;    equal,  40;    geometric,  39;    integral,  96;    of 

revolution,  234;  significant,  21,  96;  stepped,  456;   flux,  127,  182, 

of  armatures,  195 


INDEX  641 

Focal  distance,  342 

Focus,  342 

Force,  acting  on  conductor,  193;  acting  on  lever  arm,  261;  concurrent, 
532 

Formula,  48,  124;  analysis  of,  124;  capacity,  516;  conversion,  398; 
data  of,  124;  decaying  current,  481;  inductance,  509;  integra- 
tion area,  464;  interpretation,  131;  moment  of  inertia,  470; 
notation  of,  124;  solution  of,  125;  transformation  of,  125 

Formulation,  of  laws,  74,  127;  of  graphs,  348 

Fraction,  axiom,  76;  compounds,  81;  improper,  21;  operations, 
75-82;  proper,  20;  value  of,  75,  76 

Frequency,  404,  406;  comparison  of  curves  of  different,  406;  critical, 
554;  factor,  408;  standard,  546 

Frustum,  of  paraboloid,  238;  of  pyramid  or  cone,  231 

Function,  extent,  119;  for  any  quadrant,  67,  68;  multiplier  of,  119; 
of  algebraic  sums  of  angles,  396;  of  half  angles,  397;  of  multiple 
angles,  397;  symbol,  329,  330;  tangent,  397;  trigonometric,  63,  65 

Fundamental  sine  curve,  403;  principles,  A.  C.  circuit,  603 

G 

Gauge  points,  115 
Generator  curve,  374 
Generatrix,  374 
Geometric  constructions,  21 
Geometry,  36 

Graph,  282;    elements  of,  348;    formulation  of,  348,  linear,  300,  307, 
308;    of  simple  equations,  301;    solution  of  A.  C.  problem,    539 
Gravity,  center  of,  264 
Growing  current  curve,  482 
Guldinus'  theorems,  234 
Gyration,  radius  of,  268 


H 


Harmonic  motion,  401-422 
Heating  of  magnets,  200 
Helix,  217 
Henry,  506 
Hexahedron,  239 
Homogeneous  equation,  256 
Homologous  angles  and  sides,  43 
Horse-power  of  an  engine,  212 


642  INDEX 

Hyperbola,  60;    area  of  segment  curve,  211,    324;  eccentricity,  342; 

equation,  333;    length,  211;    linear  representation,  359;   odd-odi. 

odd-even,  even-odd,  320;  properties,  347 
Hyperboloid,  234,  238 
Hypocycloid,  375 
Hypotenuse,  41 
Hypothesis,  35,  38,  39 
Hysteresis,  loop,  479;   loss,  191,  323,  479 


I-beam,  270 

Icosahedron,  239 

Image,  58 

Imbedded  winding,  198 

Impedance,  522,  544,  546 

Inclination  of  vector,  525,  537 

Increment  of  distance  and  time,  434 

Indefinite  integral,  464 » 

Index,  359,  361,  383 

Indicated  operation,  254,  447 

Induced  E.M.F.,  504 

Inductance,  504;  unit  of,  506;  mutual,  509;  reactance,  521 

Inequality,  34 

Infinity  symbol,  69 

Initial  side,  27 

Input,  167,  179 

Instructions  for  preparing  and  reporting  daily  work,  631 

Integral,  430;  double,  459;  curve,  scale  of,  461;  definite,  465;  degree 
of,  equation,  461;  figures,  96;  formula  for  area,  464;  indefinite, 
464;  nomenclature  and  interpretation  of,  expressions,  475;  of 
power  of  a  variable,  447;  table  of,  450-454 

Integration,  447;  graphic,  455;  determination  of  static  moment,  472; 
moment  of  inertia  and  center  of  gravity ,  472 ;  double,  459;  graphic, 
455 

Integrand,  466 

Intercepts,  309 

Interpolation,  66,  284 

Interpretation  of  differential  and  integral  expressions,  475;  of  for- 
mulas, 131 

Inverse,  function,  428;  operation,  254 ;  setting,  110 

Involute,  377 


INDEX  643 


Joint,  efficiency,  277;  lap  and  butt,  278 
Joule,  165 


K 

Kirchhoff's  laws,  170,  171,  539 


Lag,  415 

Latitude  of  line,  525 

Law,  compound  interest,  363;  cosines,  91;  derivative  of  power,  444; 
formulation  of,  127;  numbers,  82-86;  of  duality,  284;  projective 
segments,  93;  sines,  90 

Lead,  415 

Least  common  denominator,  79 

Length  of  circular  arc,  206,  207 

Levels,  456 

Linear  coordinates,  383;  graphs,  300,  307,  308,  350;  representation 
of  equations,  352,  359-362 

Lines,  24;  axal,  367;  bisection,  31;  circularization,  208;  diagram,  383; 
non-parallel,  24;  non-perpendicular,  24;  parallel,  24;  perpen- 
dicular, 24;  reference,  280;  theorems,  35-37;  zero,  383 

Load  total,  259 

Locus,  282,  297 

Logarithm,  94-104;  Briggs  or  common,  94,  102;  definition,  95; 
hyperbolic  or  natural  or  Napierian,  102;  power,  101;  product, 
99,  100;  quotient,  100;  root,  101;  scale,  114;  second,  103,  114; 
table  of,  122, 123 

Logarithmic  curve,  291;  cross-section  paper,  351;  extrapolation,  356; 
plotting,  352;    semi,  paper,  358;    spiral,  372;    unit  square,  351 

Loop,  404 


M 

Magnetizing  force,  182 

Magnetomotive  force,  182 

Magnification  of  area,  287 

Magnitude,  1,  2,  47;  of  vector,  524,  525,  537 


644  INDEX 

Mantissa,  96 

Mass,  center  of  gravity  of,  264 

Mathematic  authorities,  35 

Matrix,  57,  58 

Maximum  point,  327,  404 

Mean,  effective  value,  463,  468;  ordinate  rule,  213;  proportional,  373 

Means,  53 

Measure,  x,  394 

Measurement,  1,  21,  22,  24;  angles,  29 

Measuring  instruments,  electrical,  495-503 

Mechanical  drawings,  58 

Median,  49 

Megohm,  130 

Member  of  equation,  253 

Microhm,  130 

Mil-foot  circular,  138,  202 

Minimum  point,  327,  404 

Minuend,  9,  10,  12 

Minus  sign,  9,  10,  11;  double,  11 

Modulus,  102;  of  elasticity,  260 

Motion,  harmonic,  401 

Multiplier,  13,  14;  of  resistance,  502 

Multiplication,  algebraic,  17,  83;    arithmetic,  13;    by  slide-rule,  107, 

108;  graphic,  293;  logarithmic,  94-99, 100;  symbol,  81 
Multisection  of  an  angle,  380 
Mutual  inductance,  509 


N 


Napierian  number,  102,  114,  363 

Nappes,  60,  61 

Natural  period,  554 

Net  work  of  circuit,  170 

Neutral  axis,  268 

Nock,  526 

Nomenclature  of  differential  and  integral  expressions,  475-482 

Non-linear,  equations,  316;  simultaneous,  337 

Notations,  75,  124,  131,  145;  of  vectors,  537 

Numbers,  4;  laws  of,  82-86;  Napierian,  363;  of  turns  of  winding,  199; 

worth,  112 
Numerator,  20 


INDEX  645 

O 

Octahedron,  239 

Odd-even  hyperbolas  and  parabolas,  320 

Odd-odd  hyperbolas  and  parabolas,  320 

Ohm's  law,  74,  124 

Ohmic  factor,  547 

Operations,  axiom,  31;   fractions,  74-82;   indicated,  254;   inverse,  254 

Ordinate,  283 

Origin,  282,  367;  shifting,  287 

Orthogonal  projection,  58 

Output,  167 


Paper,  logarithmic,  352,  354;   polar,  367;   semi-log,  358;   special,  359, 

360,  362;  squared,  22;  use  of,  279-297 
Parabola,  60;    area,  210;    chord,  210;    curve,  317;    eccentricity,  342; 

equation,    333,    334;     length,    211;     linear    representation,    351; 

odd-even,  odd-odd  and  even-odd,  320;    standard,  317,  320 
Paraboloid,  frustum,  238;   generation,  234,  238 
Parallelogram,  43;  area,  49 
Parameter,  320 
Parenthesis,  12 
Parts,  40 
Per  cent,  48 
Period,  406 

Permeability,  183,  185 
Permeance,  183;  leakage,  188-190 
Perpendicular,  24,  36-39,  62 
Phase,  413,  415;  construction  of,  angle,  607 
x  measure,  394 
Pictorial  representation,  279 
Place,  279 
Planimeter,  212 
Plate,  300 
Plotting  first  degree  equation,  311;   of  formulas,  295;   on  logarithmic 

paper,  352;    on  squared  paper,  353;    third  variable,  290;    more 

than  three  variables,  390 
Plus  sign,  4,  7,  10,  11 
Point,  35-37;    coordinates,  283,  351,  367;    line  axiom,  35;    maximum 

and  minimum,  327 


646  INDEX 

Polar,  coordinate,  367;    equation  of  circle,  371;    moment  of  inertia, 

274;  paper,  use  of,  367 
Pole,  342,  367;  unit,  457 
Polygon,  40,  44,  45;  area,  regular,  204 
Polynominal,  82 
Power,  162;    axiom,  32;    by  a  curve,  324;    by  logarithms,  101;    by 

log  plotting,  356;    by  slide-rule,  113;    diameter  of  wire  for,  177; 

factor,  546;  law  of  derivative  of,  444;  series,  329 
Practical  values  of,  inductance,  511 ;  capacity,  515 
Primary  of  transformer,  590 
Primitive,  440;  table  of,  450-454 
Prism,  219;  decomposition,  228 
Prismoid,  volume,  232 

Product,  14, 19,  48,  83;  slide  rule,  116;  sine  curves,  423,  427 
Progression,  arithmetic,  105;  geometric,  106 
Projection,  56,  58-62;    central,  58;    geometric,  58;    of  vector,  528; 

orthogonal,  58 
Projector,  62 

Proof,  geometric,  38;  derivative  of  power,  444 

Properties,  of  geometric  constructions,  21;  ellipses  and  hyperbolas,  347 
Proportion,   51-55;    by  slide-rule,    109;    continued,   52;    direct,   57; 

factors,  55;  indirect,  57;  theorems,  54-55 
Proportionate  setting,  109 
Proportional  parts,  99 
Protractor,  23,  29 
Pulley  diameter,  276 

Pyramid,  equivalent,  229;    frustum,  231;    surface,  230;    volume.  227 
Pythagorean  theorem,  72,  244 

Q 

Quadrant,  28,  29,  63;  signs,  69 

Quadratic   equation,   247-258;    graphic   construction   of  roots,   251; 

solved    by    factoring,     248;      solved    by    completing    trinominal 

square,  249 
Quadrilateral,  40,  43;  area,  50 
Quantity,  1;  parenthetical,  257;  scalar,  524;  unknown,  247;  vector, 

524 
Quotient,  19,  20,  71 

R 

Radian  of  angle  and  arc,  29;  table  of,  121 
Radical  symbol,  13,  253 


INDEX  647 

Radius,  28;  of  gyration,  268;  vector,  27, 63,  367,  373 

Rate,  430 

Ratio,  20,  47,  51,  52,  342 

Reactance,  capacity,  522;  inductive,  508,  522;  total,  523,  543;  voltage, 

543 
Reactions,  261 
Reactors,  511,  535 
Reciprocal,   81,    113;    charts,  388;    curves,   324;    log  plotting,   356; 

ohmic  factor,  547 
Record,  of  harmonic  motion,  402 
Rectangle,  46 

Regular  polygon,  44;  area,  204 
Reluctance,  182,  185 
Reluctivity,  183 
Remainder,  9 

Representation  pictorial,  279;  linear,  of  equations,  352-362 
Resistance,   136,  138;    mnemonics,  139;    specific,  163,  183;    table  of 

copper,  629;  table  of  mil-foot,  139 
Resistivity,  183 
Resistor,  152 
Resonance,  554-555 
Resultant,  417;    continued  or  partial,  530;    resistive-reactive  circuit, 

539-545 
Reversibility  of  mathematic  processes,  284 
Revolution,  figure  of,  234 
Right,  angle,  24;  triangle,  23,  70,  71 
Ring,  solid,  circular  and  rectangular,  235 
Risers,  456 
Riveted  joints,  276 
Root,  axiom,  32;  by  a  curve,  324;  by  logarithms,  101;  by  log  plotting, 

356;   by  slide-rule,  113;   index,  101;   square  root,  arithmetic,  73; 

symbol,  13 
Rotating  body,  velocity  of,  436 
Roulettes,  374 

Rule,  of  slide-rule,  107;  flexible,  280  ' 
Runner,  111 


Scale,  1 ;  of  integral  curves,  461 
Secant,  65 

Second  derivative,  459 
Secondary  of  transformer,  590 


648  INDEX 

Section,  60;  conic,  331 

Sector  of  circle,  209 

Segment,  law  of  projective,  93;  of  circle,  209;  of  line,  43,  62;  of 
sphere,  237 

Self-induction,  477,  504-511 

Semi-log  paper,  358 

Sense  of,  inequality,  34;  vector,  525 

Series,  104;  A.  C.  circuit,  546-558;  power,  330;  parallel  circuits, 
570-572 

Setting,  diagrammatic,  109;  direct,  110;  inverse,  110;  proportional,  109 

Shadow,  £8;  line,  62 

Shaft,  round,  275 

Shunt,  galvanometer,  161;  ammeter,  502 

Significant  figure,  96 

Signs  of  quadrants,  69 

Simple  equation,  247 

Simpson's  rule,  215,  216 

Simultaneous  equations,  128,  312-315,  336-341,  349 

Sine,  64;  curve,  294;  hyperbolic,  366,  399;  law  of,  90;  linear  repre- 
sentation of,  362;  logarithmic  scale,  114;  record  of  harmonic 
motion,  402;  series,  398;  table  of,  121 

Slide-rule,  104-120 

Slope,  284,  302;  of  curve,  432;  of  linear  equation,  310 

Solenoid,- 184 

Solid  ring,  circular,  and  rectangular  sections,  235 

Solidus,  12 

Solution  of  formulas,  125,  graphic,  383 

Specific;  conductance,  183;  permeance,  183;  reluctance,  183;  resist- 
ance, 163,  183 

Sphere,  generation,  234;  oblate  and  prolate,  237;  segment,  237; 
volume,  236 

Spherical  zone,  volume,  237 

Spiral,  conical,  217;  equiangular,  372;  hyperbolic,  371;  logarithmic, 
372;  parabolic,  371;  plane,  217;  standard  linear  or  Archimedian, 
370 

Speed,  430 

Spread,  59,  61,  233 

Square,  46;  completing,  249;  mil,  140;  root,  73;  root  by  log  paper, 
356;  unit,  202;  winding,  197 

Squared  paper,  use  of,  279-299;  for  logarthmic  plotting,  351;  power 
and  root  by,  356;  use  of  for  area,  213 

Squaring  of  circle,  205 

Staff,  107 


INDEX  649 

Stagger  winding,  198 

Standard,  hyperbola,  324;  linear  spiral,  369;  parabola,  318 

Static  formula,  472 

Statistics,  279 

Stepped  figure,  456 

Stiffness  of  beams,  273 

Stock,  107 

Straightedge,  21,  23 

Strength,  comparative  of  beams,  271,  272;  of  current,  131;  materials, 
260;  table  of  materials,  262;  ultimate,  260 

Stress,  intensity,  259 

Strip  elementary,  464 

Subscript,  52 

Subtraction,  arthmetic,  9-12;  algebraic,  10-12;  of  curves,  323 

Subtrahend,  9-12 

Sum,  4,  7-9;  vector,  526-532 

Summation  symbol,  266,  470 

Support,  383 

Surface,  cone,  233;    cylinder,  233;    lateral,  60,  218;    prismatic  figure, 
227,230;  resisting,  259 

Susceptance,  545,  561 

Symbol,  abbreviation,  6;  aggregation,  12;  correspondence,  98,  297 
differentiation,  438;  equality,  4;  factorial,  240;  function,  329 
inequality,  34;  infinity,  69;  integral,  447;  multiplication,  81 
summation,  266,  470;  variation,  52;  X,  Y,  Z,  546 

Symmetry,  323 


T-square,  23 

Tangent,  64;  formula,  397;  logarithmic  scale,  114;  table  of,  121 

Temperature  coefficient,  163,  165;  table  for  copper,  299 

Tension,  259 

Term,  12;  absolute,  247;  degree,  306;  meaning  of  xy,  335 

Terminal  side,  27 

Tetrahedron,  238 

Theorem,   35-40;    binominal,  240;    Guldinus',   234;     on  limits,   227; 

Pythagorean,  72,  244 
Time,  constant  of  circuits,  507,  515;   rate  of  change  of  distance,  435; 

table,  431 
Tongue,  107 
Torque,  194 
Total  load,  259 


650  INDEX 

Transformer,  590-600;  design,  593;  losses  and  efficiency,  592;  primary, 

590;  secondary,  590;  types,  590 
Transformation,  of  axes,  286;  of  formulas,  125,  129 
Transversal,  26 
Trapezoid,  43,  49 
Trapezoidal  rule,  215 
Triangle,  40;     area,  49;    equal,  41;    right,  41;     solution  of  oblique, 

90-93;  solution  of  right,  70,  71 
Triangular  unit,  202 

Trigonometric,  function,  63;  series,  398;  table,  121;  use  of,  tables,  65 
Trinominal,  82 
Triple  integration,  474 
Trisection  of  an  angle,  377-382 
Trough,  327 
Truncated  cylinder,  59 

U 

Ungula,  232 

Unit,  board  foot,  202;  circle,  77;  circular,  202;  cubic,  202;  linear,  47; 

mil-foot,  202;  of  area,  46;  of  measure,   1,  2;  spheric,  202;  square, 

202;   triangular,  202 
Unknown  quantity,  247 
Use,  of  letters,  6;  of  trigonometric  tables,  65 


Variable,  279;  dependent,  295,  302;  independent,  295,  302;  plotting 
third,  290;  relation  of  two,  279 

Variation,  51,  279;  problems,  136;  symbol,  52 

Vector,  524-538;  addition,  526;  algebra,  536-538;  direction,  524; 
equal,  526;  inclination,  537;  magnitude,  524,  537;  representa- 
tion, 525;  resultant,  527,  538;  sense,  525;  symbolic  representa- 
tion, 536 

Velocity,  430,  435;  angular,  401,  404,  436;  average,  434;  final,  434; 
initial,  434;    linear,  436;  scale  of,  433 

Vertex  of,  triangle,  23;  conic,  323,  344 

Vinculum,  12 

Volt,  193 

Volume  cone,  229;  cylinder,  227;  cylindric  boiler,  243;  frustum,  231; 
governor  ball,  243;  hyperboloid,  238;  irregular  solid,  239;  para- 
boloid, 238;  prism,  220;  prismoid,  232;  pyramid,  227;  regular 
solid,  238;  segment  of  sphere,  237;  segment  of  spheroid,  238; 
sphere,  236,  spherical  zone,  237;  stream,  242 


INDEX  651 

W 


Watt,  165;  second,  165 

Weddle's  rule,  216 

Wedge,  volume,  232 

Weighing,  area,  213 

Winding,  imbedded,  198;  square,  197;  stagger,  193 

Work-book,  631 

Worth,  110 

Writing  the  equation  of  a,  curve,  348-366;  line,  311 


X 

X  factor  in  A.  C.  circuit,  508,  522,  523,  543,  546 

Y 

Y-connection  in  A.  C.  circuit,  598 

Z 

Zone,  volume  of  spherical,  237 


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